An inequality $16(ab + ac + ad + bc + bd + cd) \le 5(a + b + c + d) + 16(abc + abd + acd + bcd)$ Let $a, b, c, d \ge 0$ be nonnegative real numbers such that $a + b + c + d \le 1$.
Show that $16(ab + ac + ad + bc + bd + cd) \le 5(a + b + c + d) + 16(abc + abd + acd + bcd)$.
If I set the power means as $A := (a + b + c + d)/4$ and $Q := \sqrt{(a^2 + b^2 + c^2 + d^2)/4}$ and $C := \sqrt[3]{(a^3 + b^3 + c^3 + d^3)/4}$, then
$$RHS - LHS = \frac{4}{3}(128A^3 - 96A^2 - 96QA + 15A + 16C^3 + 24Q^2),$$
where $0 \le A \le 1/4$. The power mean inquality states that $A \le Q \le C$. What can I do with the $-96QA$ term?

Let $F(a, b, c,d)$ denote $RHS - LHS$. Then, using the notation $\sum_{\mathrm{sym}}f(a,b ,c, d) := \frac{1}{4!} \sum_{\pi \in S_4} f(\pi(a), \pi(b), \pi(c), \pi(d))$ where $\pi$ ranges over all permutations of $\{a, b, c, d\}$,
\begin{align}
&F(a, b, c, d) - F(\frac{a + b + c + d}{4}, \frac{a + b + c + d}{4}, \frac{a + b + c + d}{4}, \frac{a + b + c + d}{4})\\
&=  24\sum_{\mathrm{sym}} a^2 - 24\sum_{\mathrm{sym}} ab + 40 \sum_{\mathrm{sym}} abc - 4 \sum_{\mathrm{sym}} a^3 - 36 \sum_{\mathrm{sym}} a^2b
\end{align}
Now $\sum_{\mathrm{sym}} a^2 - \sum_{\mathrm{sym}} ab = \sum_{\mathrm{sym}} (a^2 + b^2)/2 - \sum_{\mathrm{sym}} ab = \sum_{\mathrm{sym}} (a - b)^2/2 \ge 0$, so using $1 \ge a + b + c + d$,
\begin{align}
\sum_{\mathrm{sym}} a^2 - \sum_{\mathrm{sym}} ab 
&\ge  (a + b + c + d)\sum_{\mathrm{sym}} \frac{(a - b)^2}{2} \\
&=\sum_{\mathrm{sym}}a^3 + \sum_{\mathrm{sym}}a^2b - 2\sum_{\mathrm{sym}}abc.
\end{align}
Substituting this latter inequality to the original difference, we get
\begin{align}
&F(a, b, c, d) - F(\frac{a + b + c + d}{4}, \frac{a + b + c + d}{4}, \frac{a + b + c + d}{4}, \frac{a + b + c + d}{4})\\
&\ge 24\left( \sum_{\mathrm{sym}}a^3 +  \sum_{\mathrm{sym}}a^2b - 2\sum_{\mathrm{sym}}abc\right) + \left( 40 \sum_{\mathrm{sym}} abc - 4 \sum_{\mathrm{sym}} a^3 - 36 \sum_{\mathrm{sym}} a^2b\right) \\
&= 20 \sum_{\mathrm{sym}}a^3 - 8 \sum_{\mathrm{sym}}abc - 12\sum_{\mathrm{sym}}a^2b
\ge 0,
\end{align}
where the last inequality follows from Muirhead/AM-GM, since $(3, 0, 0, 0)$ majorises both $(2, 1, 0, 0)$ and $(1, 1, 1, 0)$.
The result then follows by noting that
\begin{align}
&F(\frac{a + b + c + d}{4}, \frac{a + b + c + d}{4}, \frac{a + b + c + d}{4}, \frac{a + b + c + d}{4}) \\
&= (a + b + c + d)( 4 - (a + b + c + d))(5 - (a + b + c + d)) \ge 0,
\end{align}
for $0 \le a + b + c + d \le 4$.
The original idea is due to Macavity.
