<s>[INST] Proving $a^{-2}bcd+b^{-2}cda+c^{-2}dab+d^{-2}abc\geqslant a+b+c+d$ 
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?
 [/INST] 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!
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