Proving that $\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}> \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

AM-GM helps! $$\sum_{cyc}\frac{ab}{c^3}=\frac{1}{4}\sum_{cyc}\left(\frac{2ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}\right)\geq\frac{1}{4}\sum_{cyc}\left(4\sqrt[4]{\left(\frac{ab}{c^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ca}{b^3}}\right)=\sum_{cyc}\frac{1}{c}.$$ Done!

Without $cyc$ we can write the solution so: $$\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}=$$ $$=\frac{1}{4}\left(\left(\frac{2ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}\right)+\left(\frac{ab}{c^3}+\frac{2bc}{a^3}+\frac{ca}{b^3}\right)+\left(\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{2ca}{b^3}\right)\right)\geq$$ $$\geq\frac{1}{4}\left(4\sqrt[4]{\left(\frac{ab}{c^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ca}{b^3}}+4\sqrt[4]{\left(\frac{bc}{a^3}\right)^2\cdot\frac{ab}{c^3}\cdot\frac{ca}{b^3}}+4\sqrt[4]{\left(\frac{ca}{b^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ab}{c^3}}\right)=$$ $$=\frac{1}{c}+\frac{1}{a}+\frac{1}{b}.$$

The same trick gives also a proof by Holder: $$\sum_{cyc}\frac{ab}{c^3}=\sqrt[4]{\left(\sum_{cyc}\frac{ab}{c^3}\right)^2\sum_{cyc}\frac{bc}{a^3}\sum_{cyc}\frac{ca}{b^3}}\geq\sum_{cyc}\sqrt[4]{\left(\frac{ab}{c^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ca}{b^3}}=\sum_{cyc}\frac{1}{c}.$$ Turned out even a bit of shorter.


Use the rearrangement inequality. Assume without loss of generality $a\geq b\geq c>0$. Then we have $ab\geq ac\geq bc\,$ and $\,1/c^3\geq 1/b^3\geq 1/a^3$. Therefore the sorted sum-product

$$\frac{ab}{c^3}+\frac{ac}{b^3}+\frac{bc}{a^3}\geq\frac{bc}{c^3}+\frac{ab}{b^3}+\frac{ac}{a^3}=\frac{b}{c^2}+\frac{a}{b^2}+\frac{c}{a^2}$$

is greater than equal to a shuffled sum-product, which is greater than equal to the reversed sum-product

$$\frac{b}{c^2}+\frac{a}{b^2}+\frac{c}{a^2}\geq\frac{c}{c^2}+\frac{b}{b^2}+\frac{a}{a^2}=\frac{1}{c}+\frac{1}{b}+\frac{1}{a}.$$

Equality is obtained if and only if $\,a=b=c$. If $a,b,c$ are different (not necessarily all different), the greater than "$>$" sign holds.

Tags:

Inequality

A.M. G.M. Inequality

Holder Inequality

Sum Of Squares Method

Rearrangement Inequality

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