Question 65
Prove that
$$
\frac{xyzt}{(x+y)(z+t)} \le \frac{(x+z)^2(y+t)^2}{4(x+y+z+t)^2}
$$
for all positive numbers $x$, $y$, $z$, $t$.
$$
\frac{xyzt}{(x+y)(z+t)} \le \frac{(x+z)^2(y+t)^2}{4(x+y+z+t)^2}
$$
for all positive numbers $x$, $y$, $z$, $t$.