<s>[INST] 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?
 [/INST] Remark: Inspired by @Macavity's answer
Let $x = a + b, \ y = c + d$.
We have
\begin{align*}
 \mathrm{RHS} - \mathrm{LHS} &= 
 5(x + y) + 16(ab y + cd x) - 16(ab + cd + xy)\\
 &= 5(x + y) - 16 xy - 16(1 - y)ab - 16(1 - x)cd\\
 &\ge 5(x + y) - 16 xy - 16(1 - y)\cdot \frac{x^2}{4} - 16(1 - x)\cdot \frac{y^2}{4} \tag{1}\\
 &= 5(x + y) - 16xy - 4x^2 - 4y^2 + 4x^2y + 4xy^2\\
 &= 5(x + y) - 8xy - 4(x + y)^2 + 4xy(x + y)\\
 &= 5(x + y) - 4xy (2 - x - y) - 4(x + y)^2\\
 &\ge 5(x + y) - (x + y)^2(2 - x - y) - 4(x + y)^2 \tag{2}\\
 &= (x + y)(1 - x - y)(5 - x - y)\\
 &\ge 0
\end{align*}
where we have used $ab \le (a + b)^2/4 = x^2/4$
and $cd \le (c + d)^2/4 = y^2/4$ in (1),
and $4xy \le (x + y)^2$ in (2).
We are done.
</s>