# Prove $\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$

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.

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.

• 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 $n$ – Petite Etincelle Jul 31 '14 at 8:00

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.

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

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.

By AM-GM, we have:

$$\frac{a^3}{b^2} + a\ge \frac{2a^2}{b}\Rightarrow \frac{a^3}{b^2}\geq \frac{2a^2}{b} -a$$

Similar: $$\frac{b^3}{c^2}\geq \frac{2b^2}{c} -b$$

$$\frac{c^3}{a^2}\geq \frac{2c^2}{a} -c$$

Now adding them together and using C-S, we have:

$$\frac{a^3}{b^2}+\frac{b^3}{c^2} + \frac{c^3}{a^2}$$

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

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

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

Equality holds when $$a=b=c$$

P.s: I don't think it's a nice solution:(