Proofs of the Weitzenbock inequality: $a^2+b^2+c^2\geq 4 \sqrt{3}\cdot(\text{area of }\triangle ABC)$

I'm collecting proofs for Weitzenbock's inequality, once featured as a question in IMO 1961. I made three proofs for this. See these proofs below. (Whoever has a more cool proof, please share.)

Weitzenbock's inequality. Let $a$, $b$, $c$ be the sides of the triangle, and let $A$ be its area. Then:

$$a^{2}+b^{2}+c^{2}\;\geq\;4\sqrt{3}\;A$$

Proof 1

Let $R$ the circunradius. Suppose, by contradiction, that: \begin{equation} \frac{1}{a}+\frac{1}{b}+\frac{1}{c}<\frac{3}{R\sqrt{3}} \tag{1} \end{equation} Using that: \begin{equation} a+b+c\leq 3R\sqrt{3} \tag{2} \end{equation} Multiplying (1) e (2): $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)<9$$ What is absurd, this can be seen using the inequality that relates the arithmetic mean to the harmonic mean, or by the inequality of Cauchy-Schwarz. From where we conclude: \begin{equation} \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{3}{R\sqrt{3}} \end{equation} Using $\displaystyle A=\frac{abc}{4R}$, we get: $$ab+ac+bc\geq \frac{\sqrt{3}abc}{R}$$ $$ab+ac+bc\geq 4A\sqrt{3}$$ This result follows using that $\displaystyle a^{2}+b^{2}+c^{2}\geq ab+bc+ac$

Proof 2

By Heron we get:

\begin{equation*} a^2+b^2+c^2 \geq 4\sqrt{3} \sqrt{S(S-a)(S-b)(S-c)} \end{equation*}

Multiplying both sides of the inequality above by $\displaystyle \sqrt{\left(\frac{a+b+c}{2abc}\right)^2}=\sqrt{\left(\frac{S}{abc}\right)^2}$, : \begin{equation*} \left(a^2+b^2+c^2\right)\left(\frac{a+b+c}{2abc}\right) \geq 4\sqrt{3} \sqrt{\frac{S(S-a)}{bc}\frac{S(S-b)}{ac}\frac{S(S-c)}{ab}} \end{equation*} By the law of cosine and law of sine we have:

\begin{equation*} \left(a^2+b^2+c^2\right)\left(\frac{a+b+c}{2abc}\right) \geq 4\sqrt{3} \cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\gamma}{2} \end{equation*}

\begin{equation*} \left(\frac{a^2}{bc}+\frac{a}{b}+\frac{a}{c}\right)+ \left(\frac{b^2}{ac}+\frac{b}{a}+\frac{b}{c}\right)+ \left(\frac{c^2}{ab}+\frac{c}{a}+\frac{c}{b}\right) \geq 8\sqrt{3} \cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\gamma}{2} \end{equation*}

\begin{equation*} \left(\frac{sen^2\alpha}{sen\beta sen\gamma}+\frac{sen\alpha}{sen\beta}+\frac{sen\alpha}{sen\gamma}\right)+\left(\frac{sen^2\beta}{sen\alpha sen\gamma}+\frac{sen\beta}{sen\alpha}+\frac{sen\beta}{sen\gamma}\right)+\left(\frac{sen^2\gamma}{sen\alpha sen\beta}+\frac{sen\gamma}{sen\alpha}+\frac{sen\gamma}{sen\beta}\right) \geq 8\sqrt{3} \cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\gamma}{2} \end{equation*}

\begin{equation*} (sen\alpha+sen\beta+sen\gamma)\left(\frac{sen\alpha}{sen\beta sen\gamma}+\frac{sen\beta}{sen\alpha sen\gamma}+\frac{sen\gamma}{sen\alpha sen\beta}\right) \geq 8\sqrt{3} \cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\gamma}{2} \end{equation*}

\begin{equation*} \cot\alpha+\cot\beta+\cot\gamma\geq \sqrt{3} \end{equation*} And this inequality follows from the Jensen inequality.

Proof 3

Let $\displaystyle R$ the circunradius, $\displaystyle \alpha, \beta, \gamma$ the opposite acute angles to the sides of $\displaystyle a, b, c$, so:

$\\ \\ \displaystyle a^2+b^2-c^2=\frac{4Rabc(a^2+b^2-c^2)}{4Rabc}=\frac{4AR(a^2+b^2-c^2)}{abc}=4A\times\frac{2R}{c}\times\frac{a^2+b^2-c^2}{2ab}=4Acot\gamma=4A\sqrt{cot^2\gamma}=4A\sqrt{csc^2\gamma-1}=4A\sqrt{\frac{4R^2}{c^2}-1}=4A\sqrt{\frac{4a^2b^2R^2}{a^2b^2c^2}-1}=4A\sqrt{\frac{a^2b^2}{4\left(\frac{abc}{4R}\right)^2}-1}=4A\sqrt{\frac{a^2b^2}{4A^2}-1}=2\sqrt{4A^2\left(\frac{a^2b^2}{4A^2}-1\right)}=2\sqrt{a^2b^2-4A^2}\Rightarrow a^2+b^2-c^2=2\sqrt{a^2b^2-4A^2}\\ \\$

By symmetry, we get:

\begin{equation*} 2\sqrt{a^2b^2-4A^2}=a^2+b^2-c^2 \end{equation*}

\begin{equation*} 2\sqrt{a^2c^2-4A^2}=a^2+c^2-b^2 \end{equation*}

\begin{equation*} 2\sqrt{b^2c^2-4A^2}=b^2+c^2-a^2 \end{equation*}

Multiplying these equalities two by two we will have:

\begin{equation*} 4\sqrt{a^2b^2-4A^2}\sqrt{b^2c^2-4A^2}=(a^2+b^2-c^2)(b^2+c^2-a^2) \end{equation*}

\begin{equation*} 4\sqrt{a^2c^2-4A^2}\sqrt{b^2c^2-4A^2}=(a^2+c^2-b^2)(b^2+c^2-a^2) \end{equation*}

\begin{equation*} 4\sqrt{a^2b^2-4A^2}\sqrt{a^2c^2-4A^2}=(a^2+c^2-b^2)(a^2+b^2-c^2) \end{equation*}

We know that $\displaystyle m+n\geq 2\sqrt{mn}$, it folows that:

\begin{equation*} 2(a^2b^2-4A^2+b^2c^2-4A^2)\geq (a^2+b^2-c^2)(b^2+c^2-a^2) \end{equation*}

\begin{equation*} 2(a^2c^2-4A^2+b^2c^2-4A^2)\geq(a^2+c^2-b^2)(b^2+c^2-a^2) \end{equation*}

\begin{equation*} 2(a^2b^2-4A^2+a^2c^2-4A^2)\geq(a^2+c^2-b^2)(a^2+b^2-c^2) \end{equation*}

Adding all the above inequalities we will have: \begin{equation*} 4a^2b^2+4b^2c^2+4a^2c^2-16\times 3 A^2\geq (a^2+b^2-c^2)(b^2+c^2-a^2)+(a^2+c^2-b^2)(b^2+c^2-a^2)+(a^2+c^2-b^2)(b^2+c^2-a^2) \end{equation*}

Let $ b^2+c^2-a^2=x,y=a^2+c^2-b^2,z=a^2+b^2-c^2$.Observe that

$$ 2a^2=(a^2+c^2-b^2)+(a^2+b^2-c^2)=y+z$$

$$ 2b^2=(a^2+b^2-c^2)+(b^2+c^2-a^2)=x+z$$

$$ 2c^2=(a^2+c^2-b^2)+(b^2+c^2-a^2)=x+y$$

We get:

$$ 4a^2b^2=(y+z)(x+z)$$

$$ 4a^2c^2=(y+z)(x+y)$$

$$ 4b^2c^2=(x+z)(x+y)$$

Replacing in the inequality above, we have:

\begin{equation*} (x+z)(x+y)+(y+z)(x+y)+(y+z)(x+z)-16\times 3 A^2\geq xy+xz+yz \end{equation*}

\begin{equation*} x^2+y^2+z^2+2xy+2xz+2yz-16\times 3 A^2\geq 0 \end{equation*}

\begin{equation*} (x+y+z)^2-16\times 3 A^2\geq 0 \end{equation*}

\begin{equation*} x+y+z\geq 4\sqrt{3}A \end{equation*}

Done

What about a proof (almost) without words?

Let us consider a triangle $ABC$ and its Fermat-Torricelli-Steiner point, such that $\widehat{CFA}=\widehat{AFC}=\widehat{CFB}=120^\circ$. It is well-known that $F$ lies on each line joining a vertex of $ABC$ with the free vertex of the equilateral triangle built on the opposite side. By replicating the (colored) triangles $AFC,CFB,BFA$ four times we reach the configuration depicted above. Due to the triangular holes $$ [ABV_C]+[ACV_B]+[BCV_A]\geq 3[ABC] $$ that is equivalent to the claim.

There also is an efficient variational approach. If $ABC$ is a triangle in the plane with $AB=c,AC=b$, the locus of points $P$ such that $PB^2+PC^2=b^2+c^2$ is a circle through $A$ centered at the midpoint of $BC$. It follows that by moving the vertex $A$ along such circle, till meeting the perpendicular bisector of $BC$, the area of $ABC$ increases but $AB^2+BC^2+AC^2$ stays the same. It follows that for a fixed $a^2+b^2+c^2$, the maximum area is achieved only by the equilateral triangle, and $a^2+b^2+c^2\geq 4\sqrt{3}\Delta$ follows.

Yet another way is to exploit the Cauchy-Schwarz inequality. We have $$ (1+1+1)(a^4+b^4+c^4)\geq (a^2+b^2+c^2)^2 $$ hence: $$ 4\Delta = \sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)} \leq \frac{1}{\sqrt{3}}(a^2+b^2+c^2). $$

Update. Yet another way is to prove a stronger inequality, namely $$ ab+ac+bc \geq 4\Delta\sqrt{3} $$ which follows from $ab=\frac{2\Delta}{\sin C}$ and the convexity of $\frac{1}{\sin\theta}$ over $(0,\pi)$.

There are some proofs at https://artofproblemsolving.com/community/c6h14388 and a generalization at https://artofproblemsolving.com/community/c6h2958 . Let me actually reproduce my post in the latter thread in full here (with minor modifications -- neither my exposition nor my LaTeX were particularly good back in 2005), which proves several generalizations:

1. The Pedoe and Bottema inequalities

Theorem 1. Let $ABC$ be a triangle with sidelengths $a = BC$, $b = CA$, $c = AB$ and area $S$. Let $XYZ$ be a triangle with sidelengths $x = YZ$, $y = ZX$, $z = XY$ and area $T$.

(a) (Pedoe inequality) Then, we have the inequality \begin{align} a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\geq 16ST , \end{align} with equality if and only if the triangles $ABC$ and $XYZ$ are similar.

(b) (Bottema inequality) If $P$ is an arbitrary point in the plane of triangle $ABC$, then we have the inequality

\begin{align} & x\cdot AP+y\cdot BP+z\cdot CP \\ & \geq \sqrt{\dfrac{a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)}{2}+8ST} . \end{align}

(c) (Equivalent version of (a)) Let $A = \measuredangle CAB$, $B = \measuredangle ABC$, $C = \measuredangle BCA$ be the angles of triangle $ABC$. Let $X = \measuredangle ZXY$, $Y = \measuredangle XYZ$, $Z = \measuredangle YZX$ be the angles of triangle $XYZ$. Then, \begin{align} \cot A\cot Y+\cot A\cot Z+\cot B\cot Z+\cot B\cot X+\cot C\cot X+\cot C\cot Y \geq 2 . \end{align} This inequality becomes an equality if and only if the triangles $ABC$ and $XYZ$ are similar.

Proof of Theorem 1.

(a) First note that the inequality which we must prove, \begin{align} a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\geq 16ST , \end{align} is homogeneous in the sidelengths $x$, $y$, $z$ of the triangle $XYZ$ (in fact, these sidelengths occur in the power $2$ on the left hand side, and on the right hand side they occur in the power $2$ as well, since the area of a triangle depends quadratically on its sidelengths). Hence, this inequality is invariant under any similitude transformation applied to triangle $XYZ$; in other words, we can move, reflect, rotate and stretch the triangle $XYZ$ as we wish, and the inequality remains equivalent. But, of course, by applying similitude transformations to triangle $XYZ$, we can always achieve a situation when $Y = B$ and $Z = C$ and the point $X$ lies in the same halfplane with respect to the line $BC$ as the point $A$. Hence, in order to prove Theorem 1 (a) for every two triangles $ABC$ and $XYZ$, it is enough to prove it for two triangles $ABC$ and $XYZ$ in this special situation.

So, let us WLOG assume that the triangles $ABC$ and $XYZ$ are in this special situation, i. e. that we have $Y = B$ and $Z = C$ and the point $X$ lies in the same halfplane with respect to the line $BC$ as the point $A$. We have to prove the inequality $a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\geq 16ST$.

By the cosine law in triangle $ABX$, we have

$AX^2=AB^2+XB^2-2\cdot AB\cdot XB\cdot\cos\measuredangle ABX$.

Let's figure out what this equation means. At first, $AB = c$. Then, since $B = Y$, we have $XB = XY = z$. Finally, we have either $\measuredangle ABX = \measuredangle ABC - \measuredangle XBC$ or $\measuredangle ABX = \measuredangle XBC - \measuredangle ABC$ (depending on the arrangement of the points), but in both cases $\cos \measuredangle ABX = \cos \left(\measuredangle ABC - \measuredangle XBC\right)$. Since $B = Y$ and $C = Z$, we can rewrite the angle $\measuredangle XBC$ as $\measuredangle XYZ$. Thus,

$\cos\measuredangle ABX=\cos\left(\measuredangle ABC-\measuredangle XYZ\right)=\cos\measuredangle ABC\cos\measuredangle XYZ+\sin\measuredangle ABC\sin\measuredangle XYZ$.

By the cosine law in triangles $ABC$ and $XYZ$, we have

$\cos\measuredangle ABC=\dfrac{c^2+a^2-b^2}{2ca}$ and $\cos\measuredangle XYZ=\dfrac{z^2+x^2-y^2}{2zx}$.

Also, by a well-known formula, the area $S$ of triangle $ABC$ equals $S=\dfrac12\cdot ca\sin\measuredangle ABC$; thus, $\sin\measuredangle ABC=\dfrac{2S}{ca}$. Similarly, $\sin\measuredangle XYZ=\dfrac{2T}{zx}$.

Hence,

$\cos\measuredangle ABX=\cos\measuredangle ABC\cos\measuredangle XYZ+\sin\measuredangle ABC\sin\measuredangle XYZ$ $=\dfrac{c^2+a^2-b^2}{2ca}\cdot\dfrac{z^2+x^2-y^2}{2zx}+\dfrac{2S}{ca}\cdot\dfrac{2T}{zx}$.

Together with $AB = c$ and $XB = z$, this makes the equality

$AX^2=AB^2+XB^2-2\cdot AB\cdot XB\cdot\cos\measuredangle ABX$

transform into

$AX^2=c^2+z^2-2\cdot c\cdot z\cdot\left(\dfrac{c^2+a^2-b^2}{2ca}\cdot\dfrac{z^2+x^2-y^2}{2zx}+\dfrac{2S}{ca}\cdot\dfrac{2T}{zx}\right)$.

This immediately simplifies to

$AX^2=c^2+z^2-2\left(\dfrac{\left(c^2+a^2-b^2\right)\left(z^2+x^2-y^2\right)}{4ax}+\dfrac{4ST}{ax}\right)$.

After some algebra, this becomes

$AX^2=\dfrac{\left(a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)-2Q\right)-16ST}{2ax}$,

where $Q = \left(x-a\right)\left(c^2x-z^2a\right)$.

Since $Y = B$ and $Z = C$, we have $YZ = BC$, or, in other words, $x = a$. Thus, $x - a = 0$, so that $Q = 0$. Hence, the above formula for $AX^2$ simplifies to

$AX^2=\dfrac{\left(a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\right)-16ST}{2ax}$.

Since (obviously) $AX^2\geq 0$, we thus have

$\left(a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\right)-16ST\geq 0$,

and thus $a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\geq 16ST$. So the inequality of Theorem 1 (a) is proven. Equality holds if and only if equality holds in the inequality $AX^2\geq 0$; but this is clearly the case if and only if the points $A$ and $X$ coincide, i. e. if the triangles $ABC$ and $XYZ$ are congruent. Now, of course, since the triangle $XYZ$ we are dealing with is not the initial triangle $XYZ$, but just its image under a similitude transformation, the general equality condition is that the triangles $ABC$ and $XYZ$ are similar (not necessarily being congruent). So Theorem 1 (a) is proven.

(b) In our proof of Theorem 1 (a), we WLOG assumed that $Y = B$ and $Z = C$ and the point $X$ lies in the same halfplane with respect to the line $BC$ as the point $A$. This time we will assume the same situation, but with a little modification: The point $X$ should lie not in the same halfplane with respect to the line $BC$ as the point $A$, but exactly in the other halfplane. In other words, we have $Y = B$ and $Z = C$ and the points $A$ and $X$ lie in different halfplanes with respect to the line $BC$. Then, by a calculation similar to the one we did in the proof of Theorem 1 (a), we find

$AX^2=\dfrac{\left(a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\right)+16ST}{2ax}$

(note the $+$ sign instead of the $-$). Since $x = a$, we have $ax=a^2$, and thus we can rewrite this as

$AX^2=\dfrac{\left(a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\right)+16ST}{2a^2}$.

Hence,

$a^2\cdot AX^2=\dfrac{\left(a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\right)+16ST}{2}$ ${=\dfrac{a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)}{2}+8ST}$.

Taking the square root of this, we get

$a\cdot AX=\sqrt{\dfrac{a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)}{2}+8ST}$.

Hence, the inequality which is to be proven,

$x\cdot AP+y\cdot BP+z\cdot CP\geq \sqrt{\dfrac{a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)}{2}+8ST}$,

is equivalent to $x\cdot AP+y\cdot BP+z\cdot CP\geq a\cdot AX$.

So we have to prove $x\cdot AP+y\cdot BP+z\cdot CP\geq a\cdot AX$.

Well, first, by the Ptolemy inequality for the quadrilateral $PBXC$, we have $CX\cdot BP+XB\cdot CP\geq BC\cdot PX$. Since $B = Y$ and $C = Z$, we have $CX = ZX = y$, $XB = XY = z$ and $BC = YZ = x$, and therefore this inequality rewrites as $y\cdot BP+z\cdot CP\geq x\cdot PX$. Also, by the triangle inequality in the triangle $APX$, we have $AP+PX\geq AX$. Thus,

$x\cdot AP+y\cdot BP+z\cdot CP=x\cdot AP+\left(y\cdot BP+z\cdot CP\right)\geq x\cdot AP+x\cdot PX$ $=x\cdot\left(AP+PX\right)\geq x\cdot AX=a\cdot AX$,

where we have used $x = a$ in the last step. This proves Theorem 1 (b).

(c) By the Cosine Law in triangle $ABC$, we have $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$. On the other hand, by a well-known formula for the area of a triangle, the area $S$ of triangle $ABC$ equals $S=\dfrac12bc\sin A$. Thus, $\sin A=\dfrac{2S}{bc}$. Hence,

$\cot A=\cos A / \sin A=\dfrac{b^2+c^2-a^2}{2bc} / \dfrac{2S}{bc}=\dfrac{b^2+c^2-a^2}{4S}$.

Similarly, $\cot B=\dfrac{c^2+a^2-b^2}{4S}$. Thus, $\cot A+\cot B=\dfrac{b^2+c^2-a^2}{4S}+\dfrac{c^2+a^2-b^2}{4S}=\dfrac{2c^2}{4S}=\dfrac{c^2}{2S}$.

Similarly to how we proved $\cot A=\dfrac{b^2+c^2-a^2}{4S}$, we can find that $\cot Z=\dfrac{x^2+y^2-z^2}{4T}$. Thus,

$\cot A\cot Z+\cot B\cot Z=\left(\cot A+\cot B\right)\cdot\cot Z=\dfrac{c^2}{2S}\cdot\dfrac{x^2+y^2-z^2}{4T}$ $=\dfrac{c^2\left(x^2+y^2-z^2\right)}{8ST}$.

Similarly,

$\cot B\cot X+\cot C\cot X=\dfrac{a^2\left(y^2+z^2-x^2\right)}{8ST}$ and $\cot C\cot Y+\cot A\cot Y=\dfrac{b^2\left(z^2+x^2-y^2\right)}{8ST}$.

Thus,

$\cot A\cot Y+\cot A\cot Z+\cot B\cot Z+\cot B\cot X+\cot C\cot X+\cot C\cot Y$ $=\left(\cot B\cot X+\cot C\cot X\right)+\left(\cot C\cot Y+\cot A\cot Y\right)+\left(\cot A\cot Z+\cot B\cot Z\right)$ $=\dfrac{a^2\left(y^2+z^2-x^2\right)}{8ST}+\dfrac{b^2\left(z^2+x^2-y^2\right)}{8ST}+\dfrac{c^2\left(x^2+y^2-z^2\right)}{8ST}$ $=\dfrac{a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)}{8ST}$.

But after Theorem 1 (a), we have $a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\geq 16ST$, with equality if and only if the triangles $ABC$ and $XYZ$ are similar. Hence,

$\cot A\cot Y+\cot A\cot Z+\cot B\cot Z+\cot B\cot X+\cot C\cot X+\cot C\cot Y$ $\geq\dfrac{16ST}{8ST}=2$,

with equality if and only if the triangles $ABC$ and $XYZ$ are similar. And Theorem 1 (c) is proven. $\blacksquare$

2. Some applications of Pedoe

Note that Theorem 1 (a) yields the following geometric inequality:

Corollary 2. Let $ABC$ be a triangle with sidelengths $a = BC$, $b = CA$, $c = AB$ and area $S$. Let $u$, $v$, $w$ be three reals such that the numbers $v + w$, $w + u$, $u + v$ and $vw + wu + uv$ are all nonnegative. [Of course, these conditions are fulfilled if the reals $u$, $v$, $w$ themselves are nonnegative, but also in some other cases.] Then, $ua^2+vb^2+wc^2\geq 4\sqrt{vw+wu+uv}\cdot S$.

Before proving the Corollary, we show two useful facts:

Corollary 3 (a well-known formula). Let $XYZ$ be a triangle. Let $X = \measuredangle ZXY$, $Y = \measuredangle XYZ$, $Z = \measuredangle YZX$ be the angles of this triangle. Then, \begin{align} \cot Y\cdot\cot Z+\cot Z\cdot\cot X+\cot X\cdot\cot Y=1 . \end{align}

Proof of Corollary 3. Theorem 1 (c) (applied to $XYZ$, $x$, $y$ and $z$ instead of $ABC$, $a$, $b$ and $c$) shows that $\cot X\cot Y+\cot X\cot Z+\cot Y\cot Z+\cot Y\cot X+\cot Z\cot X+\cot Z\cot Y \geq 2$, and that this inequality becomes an equality if and only if the triangles $XYZ$ and $XYZ$ are similar. Thus, of course, this inequality becomes an equality (since the triangles $XYZ$ and $XYZ$ are similar). Hence, $\cot X\cot Y+\cot X\cot Z+\cot Y\cot Z+\cot Y\cot X+\cot Z\cot X+\cot Z\cot Y = 2$. Therefore,

$2 = \cot X\cot Y+\cot X\cot Z+\cot Y\cot Z+\cot Y\cot X+\cot Z\cot X+\cot Z\cot Y$ $= 2 \left(\cot Y\cdot\cot Z+\cot Z\cdot\cot X+\cot X\cdot\cot Y\right)$.

Dividing both sides of this equality by $2$, we obtain $1 = \cot Y\cdot\cot Z+\cot Z\cdot\cot X+\cot X\cdot\cot Y$. This proves Corollary 3. $\blacksquare$

Theorem 4 (Conway substitution theorem). Let $u$, $v$, $w$ be three reals such that the numbers $v + w$, $w + u$, $u + v$ and $vw + wu + uv$ are all nonnegative. [Of course, these conditions are fulfilled if the reals $u$, $v$, $w$ themselves are nonnegative, but also in some other cases.] Then, there exists a triangle $XYZ$ with sidelengths \begin{align} x=YZ=\sqrt{v+w}, \qquad y=ZX=\sqrt{w+u}, \qquad z=XY=\sqrt{u+v}. \end{align} This triangle satisfies \begin{align} y^2+z^2-x^2=2u, \qquad z^2+x^2-y^2=2v, \qquad x^2+y^2-z^2=2w. \end{align} The area $T$ of this triangle equals \begin{align} T=\dfrac12\sqrt{vw+wu+uv} . \end{align} If $X = \measuredangle ZXY$, $Y = \measuredangle XYZ$, $Z = \measuredangle YZX$ are the angles of this triangle, then $\cot X=\dfrac{u}{2T}$, $\cot Y=\dfrac{v}{2T}$ and $\cot Z=\dfrac{w}{2T}$.

Proof of Theorem 4. Since the numbers $v + w$, $w + u$, $u + v$ are nonnegative, their square roots $\sqrt{v+w}$, $\sqrt{w+u}$, $\sqrt{u+v}$ exist, and, of course, are nonnegative as well. Now, we have $\sqrt{w+u}+\sqrt{u+v}\geq\sqrt{v+w}$, since

$\sqrt{w+u}+\sqrt{u+v}\geq\sqrt{v+w}\ \ \ \ \ \Longleftarrow\ \ \ \ \ \left(\sqrt{w+u}+\sqrt{u+v}\right)^2\geq v+w$

$\Longleftarrow\ \ \ \ \ \left(w+u\right)+\left(u+v\right)+2\sqrt{\left(w+u\right)\left(u+v\right)}\geq v+w$

$\Longleftarrow\ \ \ \ \ 2\sqrt{\left(w+u\right)\left(u+v\right)}\geq -2u$ [this is not trivial, since $u$ needs not be $\geq 0$!]

$\Longleftarrow\ \ \ \ \ \sqrt{\left(w+u\right)\left(u+v\right)}\geq-u\ \ \ \ \ \Longleftarrow\ \ \ \ \ \left(w+u\right)\left(u+v\right)\geq\left(-u\right)^2$

$\Longleftarrow\ \ \ \ \ u^2+\left(vw+wu+uv\right)\geq u^2\ \ \ \ \ \Longleftarrow\ \ \ \ \ vw+wu+uv\geq 0$.

Similarly, $\sqrt{u+v}+\sqrt{v+w}\geq\sqrt{w+u}$ and $\sqrt{v+w}+\sqrt{w+u}\geq\sqrt{u+v}$. Thus, there exists a triangle $XYZ$ with sidelengths $x=YZ=\sqrt{v+w}$, $y=ZX=\sqrt{w+u}$, $z=XY=\sqrt{u+v}$. Denote by $X = \measuredangle ZXY$, $Y = \measuredangle XYZ$, $Z = \measuredangle YZX$ the angles of this triangle. We have

$y^2+z^2-x^2=\left(\sqrt{w+u}\right)^2+\left(\sqrt{u+v}\right)^2-\left(\sqrt{v+w}\right)^2$ $=\left(w+u\right)+\left(u+v\right)-\left(v+w\right)=2u$

and similarly $z^2+x^2-y^2=2v$ and $x^2+y^2-z^2=2w$. Now, denoting the area of triangle $XYZ$ by $T$, we note that, as shown during the proof of Theorem 1 (c), we have $\cot Z=\dfrac{x^2+y^2-z^2}{4T}$, so that $\cot Z=\dfrac{2w}{4T}=\dfrac{w}{2T}$, and similarly $\cot X=\dfrac{u}{2T}$ and $\cot Y=\dfrac{v}{2T}$.

Corollary 3 yields $\cot Y\cdot\cot Z+\cot Z\cdot\cot X+\cot X\cdot\cot Y=1$. Using the relations $\cot Z=\dfrac{w}{2T}$, $\cot X=\dfrac{u}{2T}$ and $\cot Y=\dfrac{v}{2T}$, this equality becomes $\dfrac{v}{2T}\cdot\dfrac{w}{2T}+\dfrac{w}{2T}\cdot\dfrac{u}{2T}+\dfrac{u}{2T}\cdot\dfrac{v}{2T}=1$. Hence, $vw+wu+uv=4T^2$, and thus $T=\dfrac12\sqrt{4T^2}=\dfrac12\sqrt{vw+wu+uv}$. This completes the proof of Theorem 4. $\blacksquare$

Proof of Corollary 2. Construct a triangle $XYZ$ for our values of $u$, $v$, $w$ according to Theorem 4. Then, applying Theorem 1 (a) to the triangles $ABC$ and $XYZ$, we get

$a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\geq 16ST$.

Using the formulas given in Theorem 4, we can rewrite this as

$a^2\cdot 2u+b^2\cdot 2v+c^2\cdot 2w\geq 16S\cdot\dfrac12\sqrt{vw+wu+uv}$.

This simplifies to $ua^2+vb^2+wc^2\geq 4\sqrt{vw+wu+uv}\cdot S$. This proves Corollary 2. $\blacksquare$

Corollary 5 (Weitzenböck inequality). Let $ABC$ be a triangle with sidelengths $a = BC$, $b = CA$, $c = AB$ and area $S$. Then, $a^2 + b^2 + c^2 \geq 4\sqrt{3} S$.

Proof of Corollary 5. Corollary 5 follows from Corollary 2 (applied to $u=1$, $v=1$ and $w=1$). $\blacksquare$

3. Another application of Pedoe

We can translate Corollary 2 even further to obtain a purely algebraic inequality (proposed in https://artofproblemsolving.com/community/c6h122316p696333 ):

Corollary 6. Let $x$, $y$, $z$ be three nonnegative reals. Let $u$, $v$, $w$ be three reals such that the numbers $v + w$, $w + u$, $u + v$ and $vw + wu + uv$ are all nonnegative. Then, \begin{align} u\left(y+z\right)^2 + v\left(z+x\right)^2 + w\left(x+y\right)^2 \geq 4 \sqrt{\left(vw+wu+uv\right) xyz\left(x+y+z\right)} . \end{align}

Proof of Corollary 6. We have $\left(y+z\right) + \left(z+x\right) = x+y + \underbrace{2z}_{\geq 0} \geq x+y$ and similarly $\left(z+x\right) + \left(x+y\right) \geq y+z$ and $\left(x+y\right) + \left(y+z\right) \geq z+x$. Hence, the nonnegative reals $y+z$, $z+x$ and $x+y$ are the sidelengths of a triangle. In other words, there exists a triangle $ABC$ with sidelengths $BC = y+z$, $CA = z+x$ and $AB = x+y$. Consider this triangle. Let $S$ be the area of this triangle. Let $a = BC$, $b = CA$ and $c = AB$ denote the sidelengths of this triangle (so that $a = BC = y+z$ etc.), and let $s = \dfrac12\left(a+b+c\right)$ be its semiperimeter. Then, Heron's formula for the area $S$ of triangle $ABC$ yields \begin{align} S = \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} . \end{align} But we have $a = BC = y+z$ and $b = CA = z+x$ and $c = AB = x+y$. Adding these three equalities together, we obtain $a+b+c = \left(y+z\right)+\left(z+x\right)+\left(x+y\right) = 2\left(x+y+z\right)$, so that $\dfrac12\left(a+b+c\right) = x+y+z$. Hence, $s = \dfrac12\left(a+b+c\right) = x+y+z$. Subtracting the equality $a = y+z$ from this equality, we find $s-a = \left(x+y+z\right)-\left(y+z\right) = x$. Similarly, $s-b = y$ and $s-c = z$.

Now, \begin{align} S = \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} = \sqrt{\left(x+y+z\right)xyz} \end{align} (since $s = x+y+z$ and $s-a = x$ and $s-b = y$ and $s-c = z$). But Corollary 2 yields \begin{align} ua^2+vb^2+wc^2\geq 4\sqrt{vw+wu+uv}\cdot S . \end{align} In view of $a = y+z$ and $b = z+x$ and $c = x+y$ and $S = \sqrt{\left(x+y+z\right)xyz}$, we obtain \begin{align} &u\left(y+z\right)^2 + v\left(z+x\right)^2 + w\left(x+y\right)^2 \\ &\geq 4\sqrt{vw+wu+uv}\cdot \sqrt{\left(x+y+z\right)xyz} \\ &= 4 \sqrt{\left(vw+wu+uv\right) \left(x+y+z\right) xyz} \\ &= 4 \sqrt{\left(vw+wu+uv\right) xyz\left(x+y+z\right)} . \end{align} This proves Corollary 6. $\blacksquare$

[Postscriptum: I have been atrocious with attribution and citation back in 2005. I am not sure where the ideas in the above proofs come from; I might have been just the scribe. Theorem 1 (a) was discovered in 1943 by Daniel Pedoe and much earlier (without the equality case) by Neuberg. It also appears as 10.8 in O. Bottema, R. Z. Djordjevic, R. R. Janic, D. S. Mitrinovic, P. M. Vasic, Geometric Inequalities, Wolters-Noordhoff 1969. Theorem 1 (b) appears as 12.56 in the same book and is credited to O. Bottema, Hoofdstukken uit de elementaire meetkunde, Den Haag 1944 (I don't have this book, but interestingly it is not out of print). I might have picked up the Conway substitution from posts by John Conway on the Hyacinthos newsgroup long ago.]