The problem is the following:

Prove for $a$, $b$, $c$, $d$ that


I understand the proof saying


Apply AM-GM to $ab$ and $cd$ to yield


However it then states the following:


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.


2 Answers 2


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

  • $\begingroup$ 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, 2019 at 23:19
  • 1
    $\begingroup$ Appreciate it, I see how AM-GM was applied to the second step now. Thanks! $\endgroup$
    – TG173
    Nov 3, 2019 at 6:50

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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