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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.

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Because by AM-GM and C-S we obtain: $$\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}=\sqrt[3]{\frac{ab(c+d)+cd(a+b)}{4}}\leq$$ $$\leq\sqrt[3]{\frac{\left(\frac{a+b}{2}\right)^2(c+d)+\left(\frac{c+d}{2}\right)^2(a+b)}{4}}=\sqrt[3]{\frac{a+b}{2}\cdot\frac{c+d}{2}\cdot\frac{a+b+c+d}{4}}\leq$$ $$\leq\sqrt[3]{\left(\frac{\frac{a+b}{2}+\frac{c+d}{2}+\frac{a+b+c+d}{4}}{3}\right)^3}=\frac{a+b+c+d}{4}\leq$$ $$\leq\frac{\sqrt{(1^2+1^2+1^2+1^2)(a^2+b^2+c^2+d^2)}}{4}=\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}.$$

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  • $\begingroup$ In the second step of the proof I am not sure how the 2nd fraction has a numerator $(b+c)$. When I tried factoring out $(a+b)(c+d)$ from the numerator right after applying AM-GM, I instead got $(c+d)$ in my second fraction not $(b+c)$. $\endgroup$ – James Eade Nov 2 '19 at 23:19
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    $\begingroup$ Appreciate it, I see how AM-GM was applied to the second step now. Thanks! $\endgroup$ – TG173 Nov 3 '19 at 6:50
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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 $$ \frac{a+b+c+d}{4}\leq\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}.$$

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