Problem 1452

To prove the inequality is equivalent to
$$\frac{3}{4} \leqslant \frac{y z}{x+y z}+\frac{z x}{y+z x}+\frac{x y}{z+x y}<1$$

First, prove the right inequality:
$$\begin{array}{l} \frac{y z}{x+y z}+\frac{z x}{y+z x}+\frac{x y}{z+x y} \\ =\frac{y z}{x(x+y+z)+y z}+\frac{z x}{y(x+y+z)+z x}+ \\ \frac{x y}{z(x+y+z)+x y}=\frac{y z}{x y+y z+z x+x^{2}}+ \\ \frac{z x}{x y+y z+z x+y^{2}}+\frac{x y}{x y+y z+z x+z^{2}}<\frac{y z}{x y+y z+z x} \\ +\frac{z x}{x y+y z+z x}+\frac{x y}{x y+y z+z x}=1 \end{array}$$

Next, prove the left inequality:
$$\begin{array}{l} \text { Because } \frac{y z}{x+y z}+\frac{z x}{y+z x}+\frac{x y}{z+x y} \\ =\frac{y z}{x(x+y+z)+y z}+\frac{z x}{y(x+y+z)+z x}+ \\ \frac{x y}{z(x+y+z)+x y}=\frac{y z}{(x+y)(x+z)}+ \\ \frac{z x}{(y+z)(y+x)}+\frac{x y}{(z+x)(z+y)} \end{array}$$

Therefore, the right inequality is equivalent to
$$\begin{array}{l} \frac{y z}{(x+y)(x+z)}+\frac{z x}{(y+z)(y+x)}+ \\ \frac{x y}{(z+x)(z+y)} \geqslant \frac{3}{4}, \\ \quad \text { which is equivalent to } 4 y z(y+z)+4 z x(z+x)+4 x y(x+y) \geqslant \\ 3(x+y)(y+z)(z+x), \\ \quad \text { which is equivalent to } 4 y^{2} z+4 y z^{2}+4 z^{2} x+4 z x^{2}+4 x^{2} y+ \\ 4 x y^{2} \\ \quad \geqslant 3\left(x y^{2}+y z^{2}+z x^{2}\right)+3\left(x^{2} y+y^{2} z+z^{2} x\right)+ \end{array}$$
$6 x y z$,
which is equivalent to $\left(x y^{2}+y z^{2}+z x^{2}\right)+\left(x^{2} y+y^{2} z+z^{2} x\right) \geqslant$ $6 x y z . \quad(*)$

By the 3-variable AM-GM inequality, we have
$$x y^{2}+y z^{2}+z x^{2} \geqslant 3 x y z, x^{2} y+y^{2} z+z^{2} x \geqslant 3 x y z$$

Therefore, inequality $(*)$ holds, and the original inequality is proved.