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Let $a,b,c\in{\mathbb{R^+}}$. Prove that

$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}$$

I tried to expand both, but I did not get anything useful.

user2345215
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Satvik Mashkaria
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2 Answers2

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On expanding, this is the same as showing $$\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2} \right) +\left(\frac{b}a+\frac{a}c+\frac{c}b\right)\ge \left(\frac{a}b+\frac{b}c+\frac{c}a\right)+3$$ which follows using a series of AM-GMs. Specifically: $$\begin{align} LHS &\ge \left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2} \right) +3 \\ &= \sum_{cyc}\left(\frac{a^2}{b^2}+1 \right) \ge \sum_{cyc}2\frac{a}b \\ &\ge \sum_{cyc} \frac{a}b+3 = RHS \end{align}$$

Macavity
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As Macavity says, this is the same as $$ \left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2} \right) +\left(\frac{b}a+\frac{a}c+\frac{c}b\right)\ge \left(\frac{a}b+\frac{b}c+\frac{c}a\right)+3 $$ which is the same after rearrangement as $$ \underbrace{\left(\frac ab-1\right)^2+\left(\frac bc-1\right)^2+\left(\frac ca-1\right)^2}_{\ge0} \ge\underbrace{6-\left(\frac ab+\frac bc+\frac ca\right)-\left(\frac ba+\frac ac+\frac cb\right)}_{\le0} $$ since $x+\frac1x\ge2$.

robjohn
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