Problem 1572

Prove that since $(x+y)(y+z)(z+x) \leqslant\left[\frac{2(x+y+z)}{3}\right]^{3}=\frac{8}{27}$, we only need to prove a slightly stronger conclusion:
$$\begin{array}{l} \frac{x y}{\sqrt{x y+y z}}+\frac{y z}{\sqrt{y z+x z}}+\frac{x z}{\sqrt{x z+x y}} \leqslant \frac{3 \sqrt{3}}{4} \sqrt{(x+y)(y+z)(z+x)} \Leftrightarrow \\ f=\sqrt{\frac{x}{(x+z)(z+y)} \cdot \frac{x y}{(y+z)(z+x)}}+ \\ \sqrt{\frac{y}{(x+y)(y+z)} \cdot \frac{y z}{(z+x)(x+y)}}+ \\ \sqrt{\frac{z}{(y+z)(z+x)} \cdot \frac{z x}{(x+y)(y+z)}} \leqslant \frac{3 \sqrt{3}}{4} \end{array}$$

Since $f$ is cyclically symmetric, we can assume without loss of generality that $x=\min \{x, y, z\}$. We only need to consider two cases: (i) $x \leqslant y \leqslant z$ and (ii) $x \leqslant z \leqslant y$. Since the proofs for both cases are essentially the same, we will only prove the first case.

From $x \leqslant y \leqslant z$, we get $x y \leqslant z x \leqslant y z$, $(y+z)(z+x) \geqslant(y+z)(x+y) \geqslant(x+$ $y)(z+x)$, thus
$$\frac{x y}{(y+z)(z+x)} \leqslant \frac{z x}{(x+y)(y+z)} \leqslant \frac{y z}{(z+x)(x+y)}$$

Also,
$$x(y+z) \leqslant y(z+x) \Rightarrow \frac{x}{(x+z)(z+y)} \leqslant \frac{y}{(x+y)(y+z)}$$

Similarly,
$$\frac{y}{(x+y)(y+z)} \leqslant \frac{z}{(y+z)(z+x)}$$

Thus,
$$\frac{x}{(x+z)(z+y)} \leqslant \frac{y}{(x+y)(y+z)} \leqslant \frac{z}{(y+z)(z+x)}$$

By (1), (2), and the rearrangement inequality, we have
$$\begin{array}{l} f \leqslant \sqrt{\frac{x^{2} y}{(x+y)(z+x)^{2}(y+z)}}+\sqrt{\frac{x y z}{(x+y)^{2}(y+z)^{2}}}+ \\ \sqrt{\frac{y z^{2}}{(z+x)^{2}(x+y)(y+z)}}= \\ \sqrt{\frac{x y z}{(x+y)^{2}(y+z)^{2}}}+\sqrt{\frac{y}{(x+y)(y+z)}}\left(\frac{x}{z+x}+\frac{z}{z+x}\right)= \\ \sqrt{\frac{x y z}{(x+y)^{2}(y+z)^{2}}}+2 \times \frac{1}{2} \sqrt{\frac{y}{(x+y)(y+z)}} \leqslant \\ \sqrt{3\left(\frac{x y z}{(x+y)^{2}(y+z)^{2}}+2 \times \frac{1}{4} \frac{y}{(x+y)(y+z)}\right)} \end{array}$$

Therefore, to prove $f \leqslant \frac{3 \sqrt{3}}{4}$, we only need to prove
$$\begin{array}{l} \frac{x y z}{(x+y)^{2}(y+z)^{2}}+\frac{1}{2} \times \frac{y}{(x+y)(y+z)} \leqslant \frac{9}{16} \Leftrightarrow \\ 16 x y z+8 y(x+y)(y+z) \leqslant 9(x+y)^{2}(y+z)^{2} \Leftrightarrow \\ 9 x^{2} z^{2}+y^{2} \geqslant 6 x y z \Leftrightarrow(3 x z-y)^{2} \geqslant 0 \end{array}$$

From the proof of this example, we can obtain the solution to the following competition problem:
Let $a, b, c>0$, and $a+b+c=\mathrm{F}$. Prove:
$$\frac{a}{\sqrt{\frac{1}{b}-1}}+\frac{b}{\sqrt{\frac{1}{c}-1}}+\frac{c}{\sqrt{\frac{1}{a}-1}} \leqslant \frac{3 \sqrt{3}}{4} \sqrt{(1-a)(1-b)(1-c)}$$
(2006 Mathematics and Friends Competition Problem)