Inequality 1

If a,b,c>0a,b,c> 0 , prove that syma2b3(a3+b3+c3)a2+b2+c2. \sum_{\text{sym}} \frac{a^{2}}{b} \geq \frac{3\left(a^{3}+b^{3}+c^{3}\right)}{a^{2}+b^{2}+c^{2}} .

Note by Ryad M.
1 year, 4 months ago

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Can you explain what is sym,pls.

Ferenets Roman - 1 year, 4 months ago

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It means symmetry

You can write this thing as

a2b+b2c+c2a+a2c+b2a+c2b \frac { { a }^{ 2 } }{ b }+\frac { { b }^{ 2 } }{ c } +\frac { { c }^{ 2 } }{ a }+\frac { { a }^{ 2 } }{ c }+\frac { { b }^{ 2 } }{ a } +\frac { { c }^{ 2 } }{ b }

Isaac YIU Math Studio - 1 year, 3 months ago

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It's sum over all permutation of variables

Ryad M. - 1 year ago

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Would it not be a2b+b2c+c2a\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} only!

Nikola Alfredi - 10 months ago

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@Nikola Alfredi Following it answer would be :

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

a2b3a3a2\sum \frac{a^2}{b} \geq 3\frac{\sum a^3}{\sum a^2}

First we can see that 3a39abc3\sum a^3 \geq 9abc, by AM-GM

Thus we need to show a2b×a29abc\sum \frac{a^2}{b} \times \sum a^2 \geq 9abc .

Also, byAM-GM

a2ab\sum a^2 \geq \sum ab Hence a2b×a2a2b×ab\sum \frac{a^2}{b} \times \sum a^2 \geq \sum \frac{a^2}{b} \times \sum ab

Thus by Cauchy-Schwarz Inequality :-

a2b×a2a2b×ab(a32+b32+c32)2\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 (a32+b32+c32)29abc(a^\frac{3}{2} + b^\frac{3}{2} + c^\frac{3}{2})^2 \geq 9abc by AM-GM

Thus a2b×a2a2b×ab9abc\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.

Nikola Alfredi - 10 months ago

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@Nikola Alfredi It is cyclic sum

Isaac YIU Math Studio - 10 months ago

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@Isaac Yiu Math Studio Ok, I would search about it..

Nikola Alfredi - 10 months ago

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@Nikola Alfredi You can use double cyclic sum

Ryad M. - 10 months ago

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