Problem 1495

a.
$1 - \frac{x^2 - yz}{x^2 + x} = \frac{x + yz}{x^2 + x} = \frac{1 - y - z + yz}{x^2 + x} = \frac{(1 - y)(1 - z)}{x^2 + x}$.

b.
Using a), the inequality is rewritten
$$\frac{(1-y)(1-z)}{x(x+1)} + \frac{(1-z)(1-x)}{y(y+1)} + \frac{(1-x)(1-y)}{z(z+1)} \geq 3,$$
that is
$$\frac{(x+z)(x+y)}{x[(x+z)+(x+y)]} + \frac{(y+z)(y+x)}{y[(y+z)+(y+x)]} + \frac{(z+y)(z+x)}{z[(z+y)+(z+x)]} \geq 3.$$
But applying the inequality between the arithmetic mean and the harmonic mean we deduce that
$$\frac{(x+z)(x+y)}{x[(x+z)+(x+y)]} + \frac{(y+z)(y+x)}{y[(y+z)+(y+x)]} + \frac{(z+y)(z+x)}{z[(z+y)+(z+x)]} \geq \\ \geq \frac{9}{\frac{x}{x+y} + \frac{x}{x+z} + \frac{y}{y+z} + \frac{y}{y+x} + \frac{z}{z+y} + \frac{z}{z+x}} = \frac{9}{1+1+1} = 3.$$

Alternative solution.

b.
Using a), the inequality is rewritten
$$(1-x)(1-y)(1-z) \left( \frac{1}{x-x^3} + \frac{1}{y-y^3} + \frac{1}{z-z^3} \right) \geq 3.$$
Applying the inequality between the arithmetic mean and the harmonic mean, we have
$$\begin{align*} \frac{1}{x-x^3} + \frac{1}{y-y^3} + \frac{1}{z-z^3} &\ge \frac{9}{(x+y+z)-(x^3+y^3+z^3)} \\ &= \frac{9}{1-(x^3+y^3+z^3)} \\ &= \frac{9}{(x+y+z)^3 - (x^3+y^3+z^3)} \\ &= \frac{9}{3(x+y)(y+z)(z+x)} = \frac{3}{(1-x)(1-y)(1-z)} \end{align*}$$
and the inequality is demonstrated.