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A math problem
By Qualified, history, 3 years ago, In English

Prove that for any $$$a, b, c\in \mathbb{R}^+$$$ the following inequality is true: \begin{align*} \left(\frac{a+b+c}{3}\right)\left(\frac{b^{3/2}}{\sqrt{a}}+\frac{c^{3/2}}{\sqrt{b}}+\frac{a^{3/2}}{\sqrt{c}}\right) \ \ge a(2b-a)+b(2c-b)+c(2a-c) \end{align*}

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3 years ago, # |
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Counterexample: $$$a=b=c=0.1$$$

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    3 years ago, # ^ |
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    Sorry for the inconvenience. I have fixed it. Enjoy!

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3 years ago, # |
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Auto comment: topic has been updated by Qualified (previous revision, new revision, compare).

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3 years ago, # |
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Not sure about the point of putting an MO-style inequality on codeforces, but just for the sake of completeness:

Note that by the rearrangement inequality, since $$$a^{3/2}, b^{3/2}, c^{3/2}$$$ and $$$\frac{1}{\sqrt{a}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}}$$$ are oppositely sorted, the LHS is at least $$$\frac{(a + b + c)^2}{3} \ge bc + ca + ab \ge 2(bc + ca + ab) - a^2 - b^2 - c^2$$$, where the last inequality also comes from the rearrangement inequality (or AM-GM) and we are done.

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3 years ago, # |
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You could use Holder inequality to tackle the annoying sqrt part

then the LHS become ((a+b+c)^2)/3

Then you can use almost anything to prove.(AMGM, Muirhead, etc)

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2 years ago, # |
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No need of this problem here

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2 years ago, # |
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Nice inequality

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