# Inequality – Solving $16(ab + ac + ad + bc + bd + cd) \le 5(a + b + c + d) + 16(abc + abd + acd + bcd)$

contest-mathinequality

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 $$a \in [0, \frac14]$$.