### 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*} Comments (11)
 » Counterexample: $a=b=c=0.1$
•  » » Sorry for the inconvenience. I have fixed it. Enjoy!
 » Auto comment: topic has been updated by Qualified (previous revision, new revision, compare).
 » 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.
•  » » Nice solution
 » 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)
•  » » please give me downvote i am farming downvote
 » wow qualified posting his ineqs here also