Prove that $$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$ for $a,b,c>0$
Any hints/solutions?
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Did you try expanding out the terms? – Calvin Lin Nov 17 '13 at 15:51
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Yes. Would using AM-GM on the LHS help? – user85798 Nov 17 '13 at 15:54
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Expanding the terms, we want to show that
$$ \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} + \frac{a}{c} + \frac{b}{a} + \frac{c}{b} \geq 1 + 1 + 1 + \frac{a}{b} + \frac{b}{c} + \frac{c}{a} $$
This is true because
$ \frac{a^2}{b^2} + \frac{b}{a} + \frac{b}{a} \geq 3$ and $ \frac{a^2}{b^2} + 1 \geq 2 \frac{a}{b}$ by direct AM-GM.
Add up the cyclic versions of the above inequalities.
Calvin Lin
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Also, it is equivalent to $$f\left(\frac ab\right)+f\left(\frac bc\right)+f\left(\frac ca\right)\geq 0,$$ where $$f(x)=x^2+\frac1x-x-1=\frac{(x+1)(x-1)^2}x$$ – Maximilian Janisch Jan 13 '20 at 00:16