1. ## pozitive real numbers

Let$\displaystyle x, y, z$ be pozitive real numbers satisfying $\displaystyle x+y+z =1$

$\displaystyle 6xyz \leq \frac{x(y + z)}{4 - 9yz}+\frac{y(z + x)}{4 - 9zx}+\frac{z(x + y)}{4 - 9xy}$

2. By AM–GM, we have

$\displaystyle 3\sqrt{xy}(2-3\sqrt{xy})\ \le\ \frac{3\sqrt{xy}+(2-3\sqrt{xy})}2\ =\ 1$

and

$\displaystyle 2+3\sqrt{xy}\ \le\ 2+\frac{3(x+y)}2\ =\ \frac{4+3(x+y)}2$

Also, note that $\displaystyle 2-3\sqrt{xy}$ is positive (because, by AM–GM, $\displaystyle \sqrt{xy}\le\frac{x+y}2<\frac12\,).$ So, multiplying the above two inequalities gives

$\displaystyle 3\sqrt{xy}(4-9xy)\ \le\ \frac{4+3(x+y)}2$

Hence $\displaystyle \frac{x+y}{3xy(4-9xy)}\ \ge \frac{2\sqrt{xy}}{3xy(4-9xy)}\ =\ \frac2{3\sqrt{xy}(4-9xy)}\ \ge\ \frac4{4+3(x+y)}$

Similarly, $\displaystyle \frac{y+z}{3yz(4-9yz)}\ \ge \frac4{4+3(y+z)}$ and $\displaystyle \frac{z+x}{3zx(4-9zx)}\ \ge \frac4{4+3(z+x)}.$

$\displaystyle \therefore\ \frac{x+y}{3xy(4-9xy)}+\frac{y+z}{3yz(4-9yz)}+\frac{z+x}{3zx(4-9zx)}$

$\displaystyle \ge\ \frac4{4+3(x+y)}+\frac4{4+3(y+z)}+\frac4{4+3(z+x)}$

$\displaystyle \ge\ \frac9{\frac{4+3(x+y)}4+\frac{4+3(y+z)}4+\frac{4+3 (z+x)}4}$ (by inequality of arithmetic and harmonic means)

$\displaystyle =\ \frac{36}{12+6(x+y+z)}$

$\displaystyle =\ 2$

Multiplying through by $\displaystyle 3xyz$ will give the required inequality.