# Proving $\frac{x}{\sqrt{y^2+yz+z^2}}+\frac{y}{\sqrt{z^2+zx+x^2}}+\frac{z}{\sqrt{x^2+xy+y^2}} \ge \sqrt{3}$

Let $x,y,z >0$. Prove that:$$\dfrac{x}{\sqrt{y^2+yz+z^2}}+\dfrac{y}{\sqrt{z^2+zx+x^2}}+\dfrac{z}{\sqrt{x^2+xy+y^2}} \ge \sqrt{3}$$

My solution:

By Hölder, $$\left(\sum\frac{x}{\sqrt{4y^2+yz+4z^2}}\right)^2(\sum x(4y^2+yz+4z^2)) \ge (x+y+z)^3$$

Let us denote $\sum_{sym} x^2y = X$, so we have to prove $$(x+y+z)^3 \ge 4X+3xyz \iff \sum x^3 +3xyz \ge X$$ which is Schur.

How to prove it in a different way ?

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Please avoid titles that are entirely in $\LaTeX$. –  Guess who it is. Jul 20 '12 at 11:10
I notice that you get the equality when $x=y=z$. So how about defining $\varphi(x,y,z) = \displaystyle\frac{x}{\sqrt{y^2+yz+z^2}} + \displaystyle\frac{y}{\sqrt{x^2+xz+z^2}} + \displaystyle\frac{z}{\sqrt{y^2+yx+z^2}}$, study the function $\varphi$ on $(\mathbb{R}^{+*})^3$, and prove it reaches its minimum when $x=y=z$ ? –  S4M Jul 20 '12 at 11:27

The function $f(x) = \frac 1 {\sqrt x}$ is convex, by Jensen's inequality $$\dfrac{x}{\sqrt{y^2+yz+z^2}}+\dfrac{y}{\sqrt{z^2+zx+x^2}}+\dfrac{z}{\sqrt{x^2+xy+y^2}} \geq \\ \frac {(x + y + z)^{3/2}} { \sqrt{x(y^2 + yz + z^2) + y(z^2 + zx + x^2) + z(x^2 + xy + y^2)} }$$ Our inequality becomes $$x(y^2 + yz + z^2) + y(z^2 + zx + x^2) + z(x^2 + xy + y^2) \leq \frac 1 3 (x + y + z)^3$$ Expanding both sides, it is reduced to $$xyz\leq \frac {x^3 + y^3 + z^3} {3}$$ that is true by AM-GM.