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

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

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## Comments

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TopNewestCan you explain what is sym,pls.

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It means symmetry

You can write this thing as

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

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It's sum over all permutation of variables

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

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

Also, by

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

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

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