Algebraic inequality:$\frac a{b^2} + \frac b{c^2} +\frac c{a^2} \geq \frac1a+\frac1b+ \frac1c$

Question

Prove that : If a,b,c $\in \mathbb{R^+}$

$$\frac a{b^2} + \frac b{c^2} +\frac c{a^2} \geq \frac1a+\frac1b+ \frac1c$$

My attempt :

We know that the sequence {a,b,c} and {$\frac1{a^2},\frac1{b^2},\frac1{c^2}$} are oppositely ordered thus from rearrenegement inequality we conclude -

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

Is this correct?

I think that your work by rearrangement is correct. – chloe_shi Jun 25 '14 at 05:43 — chloe_shi, Jun 25 '14 at 05:43

score 6 · Accepted Answer · edited Jul 20 '15 at 07:29

6

$\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\dfrac{a}{b^{2}}+\dfrac{b}{c^{2}}+\dfrac{c}{a^{2}}\right)\geq\left(\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a}\right)^{2}$ : Cauchy-Schwarz

$\therefore$ $\dfrac{a}{b^{2}}+\dfrac{b}{c^{2}}+\dfrac{c}{a^{2}}\geq \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$

edited Jul 20 '15 at 07:29

Daniel Fischer

206,697

answered Jun 25 '14 at 05:38

chloe_shi

2,855

Algebraic inequality:$\frac a{b^2} + \frac b{c^2} +\frac c{a^2} \geq \frac1a+\frac1b+ \frac1c$

1 Answers1