# Inequality $\frac{x^3+1}{\sqrt{x^4+y+z}}+\frac{y^3+1}{\sqrt{y^4+z+x}}+\frac{z^3+1}{\sqrt{z^4+x+y}}\geq 2\sqrt{3}$

If we Let $x,y,z>0$ such that $x+y+z=3$. how to prove that $$\frac{x^3+1}{\sqrt{x^4+y+z}}+\frac{y^3+1}{\sqrt{y^4+z+x}}+\frac{z^3+1}{\sqrt{z^4+x+y}}\geq 2\sqrt{3}$$

-
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Julian Kuelshammer Oct 25 '12 at 17:04

We shall prove the inequality with the weaker condition $x, y, z \geq 0$.

$$\sum_{cyc}{\frac{x^3+1}{\sqrt{x^4+y+z}}}=\sum_{cyc}{\frac{x^3+1}{\sqrt{x^4-x+3}}}$$

Let $f(x)={\frac{x^3+1}{\sqrt{x^4-x+3}}}$. Differentiate twice, and note that $f''(x)$ is positive for $0 \leq x \leq 1$, and that $f''(x)=0$ for exactly 1 value of $x$ between $0$ and $2$.

We shall show that the minimum must be achieved when 2 of the variables are equal. Consider 2 cases.

Case 1: 2 of the 3 variables are $\leq 1$. Then $f(x)$ is convex over $[0, 1]$, so by Jensen's inequality, the minimum must be achieved when the 2 variables are equal.

Case 2: At most 1 of the variables are $\leq 1$. Then all the variables are $\leq 2$. Since $f(x)$ is differentiable and has 1 inflection point in $[0, 2]$, by (n-1)-equal value principle, we have extrema only when 2 of the variables are equal.

(Refer to page 15 of http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf for a proof of (n-1)-equal value principle)

As such, we have shown that the minimum is achieved when 2 of the variables are equal. Set $y=x, z=3-2x$ to get the following 1 variable inequality:

$$2\left(\frac{x^3+1}{\sqrt{x^4-x+3}}\right)+\frac{(3-2x)^3+1}{\sqrt{(3-2x)^4+2x}} \geq 2 \sqrt{3}$$

Differentiating to find extrema over $[0, 1.5]$, we get $x=1$ as the only extrema. It thus suffices to check $x=0, x=1, x=1.5$.

$x=0$ gives $\frac{2}{\sqrt{3}}+\frac{28}{9}>2\sqrt{3}$. $x=1$ gives $2\sqrt{3}$. $x=1.5$ gives $2(\frac{\frac{35}{8}}{\sqrt{\frac{105}{16}}})+\frac{1}{\sqrt{3}}>2 \sqrt{3}$.

Thus $$\sum_{cyc}{\frac{x^3+1}{\sqrt{x^4+y+z}}} \geq 2\sqrt{3}$$

-