Proving $a^{-2}bcd+b^{-2}cda+c^{-2}dab+d^{-2}abc\geqslant a+b+c+d$

Question

If $a,b,c,d$ are positive real numbers, then prove that $$\frac{bcd}{a^2}+\frac{cda}{b^2}+\frac{dab}{c^2}+\frac{abc}{d^2}\geqslant a+b+c+d$$

Attempt: $$\frac{bcd}{a^2}+\frac{cda}{b^2}+\frac{dab}{c^2}+\frac{abc}{d^2} = \frac{abcd}{a^3}+\frac{abcd}{b^3}+\frac{abcd}{c^3}+\frac{abcd}{d^3}$$

Using AM-GM inequality: $$abcd\big[a^{-3}+b^{-3}+c^{-3}+d^{-3}\big]\geqslant4abcd(abcd)^{-\frac{1}{4}}$$

Could anyone help me?

Set $a=b=c=d=1$. Then you get $4>4$. – Paolo Leonetti Jul 12 '17 at 17:31 — Paolo Leonetti, Jul 12 '17 at 17:31

score 6 · Accepted Answer · answered Jul 12 '17 at 17:36

We'll replace $a\rightarrow\frac{1}{a}$ and similar.

Thuse, we need to prove that $$a^3+b^3+c^3+d^3\geq abc+abd+acd+bcd,$$ which is AM-GM or Muirhead.

For example $$\sum_{cyc}a^3=\frac{1}{3}\sum_{cyc}(a^3+b^3+c^3)\geq\frac{1}{3}\sum_{cyc}3\sqrt[3]{a^3b^3c^3}=\sum_{cyc}abc$$ and we are done!

Proving $a^{-2}bcd+b^{-2}cda+c^{-2}dab+d^{-2}abc\geqslant a+b+c+d$

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