<s>[INST] Find the minimum value of the expression $P=\frac{x^2}{\sqrt{x^3+8}}+\frac{y^2}{\sqrt{y^3+8}}+\frac{z^2}{\sqrt{z^3+8}}$ Let $x,y,z$ be positive real number such that $xy+yz+zx=3$. Find the minimum value of the expression $$P=\frac{x^2}{\sqrt{x^3+8}}+\frac{y^2}{\sqrt{y^3+8}}+\frac{z^2}{\sqrt{z^3+8}}$$
 [/INST] It is easy to guess from the symmetry that $x=y=z=1$ should give the minimum of $1$, and we will try proving this.  By Cauchy-Schwarz Inequality:
$$\left(\sum_{cyc} \frac{x^2}{\sqrt{x^3+8}} \right)\left(\sum_{cyc} \sqrt{x^3+8} \right) \ge (x+y+z)^2 $$
We will use $\sum $ to denote cyclic sums. From the above, if we show that $(x+y+z)^2 \ge \sum \sqrt{x^3+8} $, then we have $P \ge 1$.
As you noted in comments, we can use $2\sqrt{x^3+8} \le (x+2)+(x^2-2x+4) = x^2-x+6$, so it is enough to show that:
$$2\sum x^2+4\sum xy \ge \sum x^2-\sum x+18 \iff \sum x^2+\sum x \ge 6$$
But $\sum x^2 \ge \sum xy = 3$ and $\left( \sum x\right)^2 \ge 3\sum xy = 9 \implies \sum x \ge 3$, so this is true. 
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