Problem: Let $x,y,z$ be real numbers (not necessarily positive)
such that $x^4 + y^4 + z^4 + xyz = 4$.
Prove that $x \le 2$ and \[ \sqrt{2-x} \ge \frac{y+z}{2}. \]

Solution: We prove that the condition $x^4+y^4+z^4+xyz = 4$ implies
\[ \sqrt{2-x} \geq \frac{y+z}{2}. \]

We first prove the  easy part.
\begin{claim*}
  We have $x \le 2$.
\end{claim*}
\begin{proof}
  Indeed, AM-GM gives that
  \begin{align*}
    5 =x^4+y^4+(z^4+1)+xyz
      &= \frac{3x^4}{4}+\left(\frac{x^4}{4}+y^4\right)+(z^4+1)+xyz \\
      &\geq \frac{3x^4}{4}+x^2y^2+2z^2+xyz.
  \end{align*}
  We evidently have that $x^2y^2+2z^2+xyz \geq 0$ because the
  quadratic form $a^2+ab+2b^2$ is positive definite, so $x^4 \leq
  \frac{20}{3} \implies x \leq 2$.
\end{proof}
Now, the desired statement is
implied by its square, so it suffices to show that
\[ 2-x \geq \left(\frac{y+z}{2}\right)^2 \]

We are going to proceed by contradiction
(it seems that many solutions do this)
and assume that
\[ 2-x < \left( \frac{y+z}{2} \right)^2
  \iff 4x + y^2 + 2yz + z^2 > 8. \]
By AM-GM,
\begin{align*}
  x^4 + 3 & \ge 4x \\
  \tfrac{y^4+1}{2} &\ge y^2 \\
  \tfrac{z^4+1}{2} &\ge z^2
\end{align*}
which yields that
\[ x^4+\frac{y^4+z^4}{2}+2yz+4 > 8. \]
If we replace $x^4 = 4 - (y^4+z^4+xyz)$ now,
this gives
\[ -\frac{y^4+z^4}{2}+(2-x)yz > 0 \implies (2-x)yz > \frac{y^4+z^4}{2}. \]
Since $2-x$ and the right-hand side are positive,
we have $yz \geq 0$.
Now
\[ \frac{y^4+z^4}{2yz} < 2-x < \left(\frac{y+z}{2}\right)^2
  \implies 2y^4+2z^4 < yz(y+z)^2 = y^3z+2y^2z^2+yz^3. \]
This is clearly false by AM-GM, so we have a contradiction.