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A math problem
By Qualified, history, 5 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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5 years ago, hide # |
 
Vote: I like it +16 Vote: I do not like it

Counterexample:

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

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5 years ago, hide # |
 
Vote: I like it +29 Vote: I do not like it

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

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and

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are oppositely sorted, the LHS is at least

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, where the last inequality also comes from the rearrangement inequality (or AM-GM) and we are done.
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5 years ago, hide # |
 
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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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4 years ago, hide # |
 
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No need of this problem here

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4 years ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

Nice inequality

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