<s>[INST] Prove $f(-\frac12) \le \frac{3}{16}$ if all roots of $f(x) = x^4 - x^3 + a x + b$ are real 
Let $a, b$ be real numbers such that all roots of $f(x) := x^4 - x^3 + ax + b$ are real. Prove that $f(-1/2) \le 3/16$.

The question was posted recently which was closed,
due to missing of contexts etc.
My attempt: $f(-1/2) \le 3/16$ is equivalent to $a \ge 2b$.
Let $x_1, x_2, x_3, x_4$ be the real roots of $f(x)$.
By Vieta, we have
\begin{align*}
 x_1 + x_2 + x_3 + x_4 &= 1, \\
 x_1 x_2 + x_1x_3 + x_1x_4 + x_2 x_3 + x_2x_4 + x_3x_4 &= 0, \\
 x_1x_2x_3 + x_1x_2x_4 + x_1 x_3 x_4 + x_2x_3 x_4 &= -a, \\
 x_1x_2x_3x_4 &= b.
\end{align*}
The problem becomes:

If $x_1, x_2, x_3, x_4$ are real numbers such that
$x_1 + x_2 + x_3 + x_4 = 1$ and $x_1 x_2 + x_1x_3 + x_1x_4 + x_2 x_3 + x_2x_4 + x_3x_4 = 0$,
prove that
$$x_1x_2x_3 + x_1x_2x_4 + x_1 x_3 x_4 + x_2x_3 x_4 + 2x_1x_2x_3x_4 \le 0.$$

This is true (e.g. verified by Mathematica). Is there a nice proof for it (or the original problem)?
 [/INST] Suppose $b= a/2+c$ where $c>0$. Then we have
$$\underbrace{x^4-x^3+ax+{a\over 2}}_{p(x)} =-c$$
This means that graph of $p$ and line $y=-c<0$ have at least two different common points (if it has only one then $p(x)=(x-d)^4-c$ for some real $d$, this case is not possible) so $p$
has exactly two minimums with negative value, say at $x_1$ and $x_3$ where
$x_1<x_3$, so $p(x_1)<0$ and $p(x_3)<0$ i.e.
$$x_i^4-x_i^3+ax_i+{a\over 2} <0\;\;\;(*)$$
for $i=1,3$. Since $x_1,x_3$ satisfies also $p'(x_i)=0$ we have $$a=-4x_i^3+3x_i^2\;\;\;(**)$$ so we
get plugging $(**)$ in inequality $(*)$ for $i=1,3$:
$$-6x_i^4+3x_i^2<0\implies x_i^2>{1\over 2}$$
If we subtract equations $(**)$, we get
$$(x_1-x_3)\Big(4(x_1^2+x_1x_3+x_3^2)-3(x_1+x_3)\Big)=0$$
so $$4(x_1^2+x_1x_3+x_3^2)=3(x_1+x_3)$$ and from here we get
$$2(x_1^2+x_3^2)+2(x_1+x_3)^2=3(x_1+x_3)$$ which means that $$
2+2t^2<3t$$ where $t=x_1+x_3$. But this inequality does not have a
solution so we have a contradiction.
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