### Qualified's blog

By Qualified, history, 6 months ago,

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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 » 6 months ago, # |   +16 Counterexample: $a=b=c=0.1$
•  » » 6 months ago, # ^ |   0 Sorry for the inconvenience. I have fixed it. Enjoy!
 » 6 months ago, # |   0 Auto comment: topic has been updated by Qualified (previous revision, new revision, compare).
 » 6 months ago, # |   +29 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.
•  » » 6 weeks ago, # ^ |   0 Nice solution
 » 6 months ago, # |   0 You could use Holder inequality to tackle the annoying sqrt partthen the LHS become ((a+b+c)^2)/3Then you can use almost anything to prove.(AMGM, Muirhead, etc)
•  » » 6 months ago, # ^ |   -8 please give me downvote i am farming downvote
 » 6 weeks ago, # |   0 wow qualified posting his ineqs here also
 » 6 weeks ago, # |   0 Turn to AoPS please.
 » 6 weeks ago, # |   0 No need of this problem here
 » 6 weeks ago, # |   +1 Nice inequality