Prove the inequality $\sqrt{\frac{x}{y^2+z^2}}+\sqrt{\frac{y}{z^2+x^2}}+\sqrt{\frac{z}{x^2+y^2}}\ge \frac{2\sqrt{6}}3$ with $x+y+z=3,x,y,z\ge 0$

Question

Let $x\geq0$, $y\geq0$ and $z\ge 0$ such that $x+y+z=3$. Show that $$\sqrt{\dfrac{x}{y^2+z^2}}+\sqrt{\dfrac{y}{z^2+x^2}}+\sqrt{\dfrac{z}{x^2+y^2}}\ge \dfrac{2\sqrt{6}}{3}.$$

I tried C-S,Holder but without success. $$\left(\sum_{cyc}\sqrt{\dfrac{x}{y^2+z^2}}\right)^2(\sum x(y^2+z^2))\ge (x+y+z)^3$$

Answer 1

Holder helps! $$\left(\sum_{cyc}\sqrt{\frac{x}{y^2+z^2}}\right)^2\sum_{cyc}x^2(y^2+z^2)\geq(x+y+z)^3.$$ Thus, it remains to prove that $$(x+y+z)^3\geq\frac{8}{3}\sum_{cyc}x^2(y^2+z^2)$$ or $$(x+y+z)^4\geq16\sum_{cyc}x^2y^2,$$ which is true by AM-GM: $$(x+y+z)^4=\left(\sum_{cyc}(x^2+2xy)\right)^2\geq\left(2\sqrt{\sum_{cyc}x^2\cdot2\sum_{cyc}xy}\right)^2=$$ $$=8\sum_{cyc}(x^3y+x^3z+x^2yz)\geq8\sum_{cyc}(2x^2y^2)=16\sum_{cyc}x^2y^2.$$ Done!

Another way.

By AM-GM and C-S we obtain: $$\sum_{cyc}\sqrt{\frac{x}{y^2+z^2}}=\sum_{cyc}\frac{\sqrt6x}{2\sqrt{1.5x(y^2+z^2)}}\geq$$ $$\geq\sum_{cyc}\frac{\sqrt6x}{1.5x+y^2+z^2}=2\sqrt6\sum_{cyc}\frac{x^2}{3x^2+2xy^2+2xz^2}\geq$$ $$\geq\frac{2\sqrt6(x+y+z)^2}{\sum\limits_{cyc}(3x^2+2x^2y+2x^2z)}.$$ Thus, it remains to prove that $$3(x+y+z)^2\geq\sum\limits_{cyc}(3x^2+2x^2y+2x^2z)$$ or $$3(xy+xz+yz)\geq\sum_{cyc}(x^2y+x^2z)$$ or $$(x+y+z)(xy+xz+yz)\geq\sum_{cyc}(x^2y+x^2z)$$ or $$3xyz\geq0.$$ Done againe!