# Inequality 1

If $a,b,c> 0$, prove that $\sum_{\text{sym}} \frac{a^{2}}{b} \geq \frac{3\left(a^{3}+b^{3}+c^{3}\right)}{a^{2}+b^{2}+c^{2}} .$

2 years ago

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Can you explain what is sym,pls.

- 2 years ago

It means symmetry

You can write this thing as

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

- 1 year, 12 months ago

It's sum over all permutation of variables

- 1 year, 9 months ago

Would it not be $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a}$ only!

- 1 year, 6 months ago

Following it answer would be :

Proof: NOTE : All sums are symmetric in a,b,c.

$\sum \frac{a^2}{b} \geq 3\frac{\sum a^3}{\sum a^2}$

First we can see that $3\sum a^3 \geq 9abc$, by AM-GM

Thus we need to show $\sum \frac{a^2}{b} \times \sum a^2 \geq 9abc$.

Also, byAM-GM

$\sum a^2 \geq \sum ab$ Hence $\sum \frac{a^2}{b} \times \sum a^2 \geq \sum \frac{a^2}{b} \times \sum ab$

Thus by Cauchy-Schwarz Inequality :-

$\sum \frac{a^2}{b} \times \sum a^2 \geq \sum \frac{a^2}{b} \times \sum ab \geq (a^\frac{3}{2} + b^\frac{3}{2} + c^\frac{3}{2})^2$

and $(a^\frac{3}{2} + b^\frac{3}{2} + c^\frac{3}{2})^2 \geq 9abc$ by AM-GM

Thus $\sum \frac{a^2}{b} \times \sum a^2 \geq \sum \frac{a^2}{b} \times \sum ab \geq 9abc$ ...Q.E.D

If you face any problem in understanding my language or concept, feel free to ask.

- 1 year, 6 months ago

It is cyclic sum

- 1 year, 6 months ago

Ok, I would search about it..

- 1 year, 6 months ago

You can use double cyclic sum

- 1 year, 5 months ago