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."
On
This follows almost immediately by a direct application of Hölder's inequality (generalised Cauchy-Schwarz):
$$\begin{aligned} \left ( \sum \limits_{\text{cyc}} \frac{1}{y(x+y)} \right ) (y + z + x) ((x+y) + (y+z) + (z+x)) &\geq (1^{\frac{1}{3}} + 1^{\frac{1}{3}} + 1^{\frac{1}{3}})^{3}\\ \iff \sum \limits_{\text{cyc}} \frac{1}{y(x+y)} &\geq \frac{27}{(x+y+z)(2x+2y+2z)}\\ &= \frac{27}{2(x+y+z)^2}\\ \end{aligned}$$
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.