<s>[INST] Help Understanding Proof for Inequality Problem #3 from Problem Solving Strategies The problem is the following:
Prove for $a$, $b$, $c$, $d$ that
$$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$$
I understand the proof saying
$${\frac{abc+abd+acd+bcd}{4}}=\frac{(ab)(c+d)+(cd)(a+b)}{4}$$
Apply AM-GM to $ab$ and $cd$ to yield
$$\frac{\left(\frac{a+b}{2}\right)^2(c+d)+\left(\frac{c+d}{2}\right)^2(a+b)}{4}=\frac{a+b}{2}\cdot\frac{b+c}{2}\cdot\frac{a+b+c+d}{4}$$
However it then states the following:
$$\frac{a+b}{2}\cdot\frac{b+c}{2}\cdot\frac{a+b+c+d}{4}=\left(\frac{a+b+c+d}{4}\right)^3$$
I don't understand this step in the proof. Unless I'm missing something basic, the above expression does  not factor as shown. It doesn't adduce a theorem to justify the equivalence either so I don't know how it was deduced.
 [/INST] Assuming $a,b,c,d\geq 0$ Maclaurin's inequality gives
$$\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}\leq \frac{a+b+c+d}{4} $$
and by the AM-QM (i.e. Cauchy-Schwarz) inequality we have
$$ \frac{a+b+c+d}{4}\leq\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}.$$
</s>