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

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


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

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