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?
 A: 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 $a \in [0, \frac14]$.
A: 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.
A: 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.
