Proving a property of a certain function $f:[0,1] \to R$ Consider a function $f:[0,1 ]\to R $ defined by $$ f(x)=\frac{1-x}{1+x}.$$ How can we show that for $x,y \in [0,1]$:
$$ f(x)+f(y) \geq f(xy).$$ I tried but get stuck in the end.We have the inequality holds iff
$$\frac{1-x}{1+x}+\frac{1-y}{1+y} \geq \frac{1-xy}{1+xy} \geq \frac{1-xy}{(1+x)(1+y)}.$$ That will be true if $$ (1-x)(1+y)+(1-y)(1+x) \geq 1-xy.$$

Identically we have
$$
\frac{1-x}{1+x}
+
\frac{1-y}{1+y}
-
\frac{1-xy}{1+xy}
=
\frac
{(1-x)(1-y)(1-xy)}
{(1+x)(1+y)(1+xy)}
\ge
0
$$
Alternatively, continuing from where you left off
\begin{align*}
&
(1-x)(1+y)+(1-y)(1+x) \ge 1-xy
\\[4pt]
\iff\;&
2-2xy\ge 1-xy
\\[4pt]
\iff\;&
1\ge xy
\\[4pt]
\end{align*}
which is true.
