Prove $4(x + y + z)^3 > 27(x^2y + y^2z + z^2x)$

Solution 1:

This is well-know inequality, actually this is a inequality from Vacs if you are familiar with Art of Problem Solving. Here's the proof:

Let $z\le y \le x$, so we have:

$$(y-x)(y-z) \le 0 \implies y^2 - xy - zy + xz \le 0 \implies y^2 + nz \le xy + zy$$

Now for the Right Hand Side:

$$27(x^2y + y^2z + z^2x + xyz) = 27( z(y^2 + zx) + x^2y + xyz) \le 27 (z(xy + zy) + x^2y + xyz) = 27 ( 2xyz + z^2y + x^2y) = 27y(2xz + x^2 + z^2) = 27y(x+z)^2 = 4 \cdot 3^3y\left(\frac{x+z}{2}\right)\left(\frac{x+z}{2}\right)$$

Now from AM-GM we have:

$$y + \frac{x+z}{2} + \frac{x+z}{2} \ge 3 \sqrt{y\left(\frac{x+z}{2}\right)\left(\frac{x+z}{2}\right)}$$ $$(y + \frac{x+z}{2} + \frac{x+z}{2})^3 \ge 3^3 y\left(\frac{x+z}{2}\right)\left(\frac{x+z}{2}\right)$$

So we have:

$$27(x^2y + y^2z + z^x + xyz) \le 4 \cdot 3^3y\left(\frac{x+z}{2}\right)\left(\frac{x+z}{2}\right) \le (y + \frac{x+z}{2} + \frac{x+z}{2})^3 = 4(x+ y+z)^3$$

Hence the proof. As you can note this inequality is even stronger, because $xyz > 0$

Solution 2:

If $x,y,z>0$, then $$x(2x-4y-z)^2+y(2y-4z-x)^2+z(2z-4x-y)^2\gt 0$$

Rearranging gives the required inequality.

Solution 3:

Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.

Thus, by Rearrangement and AM-GM we obtain: $$x^2y+y^2z+z^2x=x\cdot xy+y\cdot yz+z\cdot zx\leq a\cdot ab+b\cdot ac+c\cdot bc=$$ $$=a^2b+abc+c^2b\leq a^2b+2abc+c^2b=b(a+c)^2=4\cdot b\left(\frac{a+c}{2}\right)^2\leq$$ $$\leq4\left(\frac{b+2\cdot\frac{a+c}{2}}{3}\right)^3=\frac{4}{27}(a+b+c)^3=\frac{4}{27}(x+y+z)^3.$$