<s>[INST] If $x,y,z\gt 0$ and $xyz=1$ Then minimum value of $\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$ 
If $x,y,z\gt 0$ and $xyz=1$ Then find the minimum value of $\displaystyle \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$

$\bf{My\; Try::}$Using Titu's Lemma $$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge \frac{(x+y+z)^2}{2(x+y+z)} = \frac{x+y+z}{2}\ge 3\frac{\sqrt[3]{xyz}}{2} = \frac{3}{2}$$
and equality holds when $$x=y=z=1$$
My question is how can we solve it without the above lemma, like using Jensen's Inequality or other inequality.
Please explain me.
Thanks
 [/INST] Use $$\frac{a^2}{b}\ge 2a-b$$
Then 
$$\frac{x^2}{y+z}=\frac14\cdot\frac{(2x)^2}{y+z}\ge\frac14(4x-y-z)$$
Similarly
$$\frac{y^2}{x+z}\ge\frac14(4y-x-z)$$
$$\frac{z^2}{x+y}\ge\frac14(4z-x-y)$$
Hence
$$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\geq \frac{1}{4}2(x+y+z) = \frac{x+y+z}{2}\geq 3\frac{\sqrt[3]{xyz}}{2} = \frac{3}{2}$$
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