USA MO 1980's inequality

Assume $x+y+z>0$, else the inequality is easy. Do some rearranging: \begin{align*}(1-x)(1-y)(1-z)\leq1-\sum\frac{x}{y+z+1}&=\sum\left(\frac{x}{x+y+z}-\frac{x}{y+z+1}\right) \\ &= \frac{1}{x+y+z}\sum\frac{x(1-x)}{y+z+1} \end{align*} Multiplying through by $x+y+z>0$, and moving everything to one side gives $$\sum x(1-x)\left[\frac{1}{y+z+1}-(1-y)(1-z)\right]\geq0.$$ Now it's just AM-GM: $$1=\frac{(1-y)+(1-z)+(y+z+1)}{3}\geq\sqrt[3]{(1-y)(1-z)(y+z+1)},$$ so $\frac{1}{y+z+1}\geq(1-y)(1-z)$, which implies the result.

Show $ \ \dfrac{x}{y+z+1}+\dfrac{y}{z+x+1}+\dfrac{z}{x+y+1} \leq 1-(1-x)(1-y)(1-z)$

WLOG, $x \geq y \geq z$

Then $ \ \dfrac{y}{z+x+1} \leq \dfrac{y}{y+z+1} \ $ and $ \ \dfrac{z}{x+y+1} \leq \dfrac{z}{y+z+1}$

So LHS $\leq \dfrac{x+y+z}{y+z+1}$ and hence it suffices to show,

$(1-x)(1-y)(1-z) \leq 1 - \dfrac{x+y+z}{y+z+1} = \dfrac{1-x}{y+z+1}$

or it suffices to show $(1-y)(1-z) \leq \dfrac{1}{(1+y)(1+z)} \ $
[as $ \ \dfrac{1}{(1+y)(1+z)} \leq \dfrac{1}{y+z+1}$]

or to show that $(1-y^2)(1-z^2) \leq 1 \ $ which is obviously true.