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?

Another way which works for several such inequalities, consider minimising $RHS - LHS$ (note there will exist a minimum for this function in this domain).  At the point of minimum, if possible let there be two unequal variables, WLOG say $a \neq b$.  Then consider
$$RHS - LHS = 16\left(ab(c+d)+cd(a+b)\right)+5(a+b+c+d) - 16\left(ab+(a+b)(c+d)+cd \right) \\= -16(1-c-d)\color{red}{ab} + (\color{red}{a+b})\left(cd+5-16(c+d)\right)+5(c+d)-16cd$$
Now it can be seen that replacing both $a, b$ with $\frac12(a+b)$ will change only the first term, and in fact it will reduce $RHS-LHS$ as $ab$ will increase.  Hence at the minimum, we must have $a=b=c=d$, so it is enough to check the inequality for this case, which is a lot easier.
$16(6a^2) \leqslant 5(4a) + 16(4a^3) \iff 4a(1-4a)(5-4a) \geqslant 0$ which is true for $a \in [0, \frac14]$.
