USA MO 1980's inequality Let $x;y;z$ be real numbers and $x;y;z \in [0;1]$
Prove: $\dfrac{x}{y+z+1}+\dfrac{y}{z+x+1}+\dfrac{z}{x+y+1} \leq 1-(1-x)(1-y)(1-z)$
The official solution uses calculus, so I can't understand it.
I hope someone have a "simpler solution" (for middle-school student). Thank you.
 A: 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.
A: 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.
