Problem: Positive real numbers $x, y, z$ satisfy $xyz+xy+yz+zx = x+y+z+1$.
Prove that
\[ \frac{1}{3} \left( \sqrt{\frac{1+x^2}{1+x}}
  + \sqrt{\frac{1+y^2}{1+y}}
  + \sqrt{\frac{1+z^2}{1+z}} \right)
  \le \left( \frac{x+y+z}{3} \right)^{5/8} . \]

Solution: The key is the identity
\begin{align*}
  \frac{x^2+1}{x+1} &= \frac{(x^2+1)(y+1)(z+1)}{(x+1)(y+1)(z+1)} \\
  &= \frac{x(xyz+xy+xz)+x^2+yz+y+z+1}{2(1+x+y+z)} \\
  &= \frac{x(x+y+z+1-yz) + x^2+yz+y+z+1}{2(1+x+y+z)} \\
  &= \frac{(x+y)(x+z) + x^2 + \left( x-xyz+y+z+1 \right)}{2(1+x+y+z)} \\
  &= \frac{2(x+y)(x+z)}{2(1+x+y+z)} \\
  &= \frac{(x+y)(x+z)}{1+x+y+z}.
\end{align*}
\begin{remark*}
  The ``trick'' can be rephrased as
  $(x^2+1)(y+1)(z+1) = 2(x+y)(x+z)$.
\end{remark*}
After this, straight Cauchy in the obvious way will do it
(reducing everything to an inequality in $s = x+y+z$).
One writes
\begin{align*}
  \left( \sum_{\text{cyc}} \frac{\sqrt{(x+y)(x+z)}}{\sqrt{1+s}} \right)^2
  &\le \frac{\left( \sum_{\text{cyc}} x+y \right)
    \left( \sum_{\text{cyc}} x+z \right)}{1+s} \\
  &= \frac{4s^2}{1+s}
\end{align*}
and so it suffices to check that $\frac{4s^2}{1+s} \le 9(s/3)^{5/4}$,
which is true because
\[ (s/3)^5 \cdot 9^4 \cdot (1+s)^4 - (4s^2)^4
  = s^5 (s-3)^2 (27s^2 + 14s + 3) \ge 0. \]