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If $a$, $b$ and $c$ are positive real numbers, prove that: $$\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$$

Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction.

Things I have tried so far:

Using Cauchy inequality I can write:$$\left(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\right)(a+b+c) \geq \left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2$$

but I can't continue this.I tried expanded form:$$\sum \limits_{cyc} \frac{a^5c^2}{a^2b^2c^2} \geq \sum \limits_{cyc} \frac{a^3c}{abc}$$

Which proceeds me to this Cauchy:$$\sum \limits_{cyc} \frac{a^5c^2}{abc}\sum \limits_{cyc}a(abc)\geq \left(\sum \limits_{cyc}a^3c\right)^2$$ I can't continue this one too.

The main Challenge is $3$ fraction on both sides which all of them have different denominator.and it seems like using Cauchy from first step won't leads to anything good.

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The next thing you can try is Cauchy again $$\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)(b+c+a)\geq (a+b+c)^2.$$ So $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c.$$ Now you can eliminate $a+b+c$ from your first inequality.

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  • 1
    $\begingroup$ You set the departure point for induction, then we can prove $\frac{a^{n+1}}{b^n} + \frac{b^{n+1}}{c^n} + \frac{c^{n+1}}{a^n} \geq \frac{a^{n}}{b^{n-1}} + \frac{b^{n}}{c^{n-1}} + \frac{c^{n}}{a^{n-1}}$ for any positive integer $n$ $\endgroup$ – Petite Etincelle Jul 31 '14 at 8:00
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Another way to do this would be the following (I'm doing Liu Gang's suggested generalization):

We have to show $$\frac{a^{n+1}}{b^n} + \frac{b^{n+1}}{c^n} + \frac{c^{n+1}}{a^n} - \frac{a^n}{b^{n-1}} - \frac{b^n}{c^{n-1}} - \frac{c^n}{a^{n-1}} \ge 0.$$ The left hand side equals $$\frac{a^n(a - b)}{b^n} + \frac{b^n(b-c)}{c^n} + \frac{c^n(c-a)}{a^n},$$ and therefore it is enough to show that $$c^n a^{2n} (a-b) + a^n b^{2n} (b-c) + b^n c^{2n} (c-a) \ge 0.$$ Because the inequality is cyclic, we can assume that either $a \ge b \ge c$ or $a \ge c \ge b$.

In the first case we have $c^n a^{2n} \ge b^n c^{2n}$ and $a^n b^{2n} \ge b^n c^{2n}$, so we get that the LHS is $\ge b^n c^{2n} (a - b + b - c + c - a) = 0$.

In the second case we have $a^n b^{2n} \le c^n a^{2n}$ and $b^n c^{2n} \le c^n a^{2n}$, so we get that the LHS is $\ge c^n a^{2n} (a - b + b - c + c- a) = 0$.

This proves the claim.

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$$\sum_{cyc}\frac{a^3}{b^2}-\sum_{cyc}\frac{a^2}{b}=\sum_{cyc}\left(\frac{a^3}{b^2}-\frac{a^2}{b}\right)=$$ $$=\sum_{cyc}\left(\frac{a^2(a-b)}{b^2}-(a-b)\right)=\sum_{cyc}\frac{(a-b)^2(a+b)}{b^2}\geq0.$$ Done!

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By AM–GM, we have $$14\frac{a^3}{b^2}+3\frac{b^3}{c^2}+2\frac{c^3}{a^2}\geq 19\frac{a^2}{b}\quad (1)$$ $$2\frac{a^3}{b^2}+14\frac{b^3}{c^2}+3\frac{c^3}{a^2}\geq 19\frac{b^2}{c}\quad (2)$$ $$3\frac{a^3}{b^2}+2\frac{b^3}{c^2}+14\frac{c^3}{a^2}\geq 19\frac{c^2}{a}\quad (3)$$ Add all three equations together and conclude.

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