Question about sines of angles in an acute triangle

Let $\triangle ABC$ be a triangle such that each angle is less than $ 90^\circ $. I want to prove that $\sin A + \sin B + \sin C > 2$.

Here is what I have done:

Since $A+B+C=180^{\circ}$ and $0 < A,B,C < 90^\circ$, at least two of $A,B,C$ are in the range 45 < x < 90, without loss of generality, let these angles be $A$ and $B$.

$\sin A + \sin B + \sin C = \sin A + \sin B + \sin(180^\circ-A-B) = \sin A + \sin B + \sin(A+B)$

Since $45^\circ < A,B < 90^\circ$ it follows that $2^{0.5} < \sin A + \sin B < 2.$ Am I near the answer?

Solution 1:

Multiplying both sides by $2R$ and exploiting the sine theorem, we have to prove that:

The perimeter of acute triangle is always greater than four times the circumradius.

Assume that $A,B$ and the circumradius $R$ are fixed. Since $ABC$ is an acute triangle, the circumcenter $O$ of $ABC$ lies inside $ABC$, so $C$ lies between the antipode $A'$ of $A$ and the antipode $B'$ of $B$ in the circumcircle $\Gamma$:

Let $M$ be the midpoint of the arc $A'B'$ and $\Gamma_C$ be the ellipse through $C$ with foci in $A,B$, i.e. the locus of points $P$ for which $PA+PB=CA+CB$. Since $A'$ and $B'$ lie inside $\Gamma_C$, we have $A'A+A'B<CA+CB$, so the perimeter of $ABC$ is minimized when $C\equiv A'$ or $C\equiv B'$, i.e. when $\widehat{C}=\frac{\pi}{2}$. This gives that the minimum perimeter is achieved in the limit case when $A$ and $B\equiv C$ are endpoints of a diameter of $\Gamma$. In such a case, obviously, the perimeter is $4R$.

As an alternative, we can use concavity. Since $\sin x$ is a concave function over $[0,\pi/2]$, $$ f(A,B,C)=\sin A+\sin B+\sin C $$ is a concave function over the set $E=\{(A,B,C)\in[0,\pi/2]^3:A+B+C=\pi\}$, so its minima lie on $\partial E$.

Solution 2:

I have found a simpler solution.

Observing the graph of $y = \sin x$ and $y = \frac{2}{\pi}x$, $x\in[0,\frac{\pi}{2}]$.
We can see that $\sin x\ge\frac{2}{\pi}x$.

Let $\bigtriangleup ABC$ be a triangle such that each angle is less than $90^{\circ}$. So we have: $A\in(0,\frac{\pi}{2}),B\in(0,\frac{\pi}{2}),C\in(0,\frac{\pi}{2})$.

Because $\sin x>\frac{2}{\pi}x$, when $x\in(0,\frac{\pi}{2})$,
$$\sin A+\sin B+\sin C > \frac{2}{\pi}(A+B+C)=2$$