<s>[INST] Interesting 3 Variable Inequality for Real Numbers I recently saw an olympiad style inequality that seemed very difficult. I tried to use elementary inequalities such as AM-GM or Cauchy-Schwarz, but neither have helped in making significant progress. Could anyone provide a rigorous proof, preferably using more elementary inequalities?
Problem: "Prove the inequality $\frac{1}{y(x+y)}+\frac{1}{z(y+z)}+\frac{1}{x(z+x)}\geq\frac{27}{2(x+y+z)^2}$ if $x,y,z$ are positive reals."
 [/INST] The inequality is homogenous in $(x, y, z)$, therefore we can assume
that $xyz=1$. Then
$$
\frac{1}{y(x+y)}+\frac{1}{z(y+z)}+\frac{1}{x(z+x)} \ge \dfrac{3}{2}
$$
as for example demonstrated in
If $xyz=1$, prove $\frac{1}{y(x+y)}+\frac{1}{z(y+z)}+\frac{1}{x(z+x)} \geqslant \frac{3}{2}$ or A inequality proposed at Zhautykov Olympiad 2008.
Also
$$
 1 = \sqrt[3]{xyz} \le \frac{x+y+z}3 \Longrightarrow
 (x+y+z)^2 \ge 9 \, .
$$
from the inequality between the geometric and the arithmetic mean.
Combining these estimates gives the desired inequality.
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