Let $\displaystyle a_1, a_2, ... , a_n$ be positive real numbers and we know that $\displaystyle a_{n+1}=a_1$

Show that:

$\displaystyle \sum_{i=1}^{n} \frac{a_i^3}{a_{i+1}^2} \ge \sum_{i=1}^{n}\frac{a_i^2}{a_{i+1}} $

I think applying Cauchy-Schwarz may be helpful but I haven't figure it out yet.

  • $\begingroup$ $a_{n+1}=a_1$? a typo? $\endgroup$ – FunctionOfX Nov 10 '14 at 12:41
  • $\begingroup$ noo that's good $\endgroup$ – John Nov 10 '14 at 12:44
  • $\begingroup$ oh, sorry, my bad $\endgroup$ – FunctionOfX Nov 10 '14 at 12:45

By Cauchy-Schwarz Inequality: $$\left(\sum_{cyc} \frac{a_i^2}{a_{i+1}}\right)^2 \leqslant \left( \sum_{cyc} a_i \right) \left( \sum_{cyc} \frac{a_i^3}{a_{i+1}^2} \right)$$

So it is enough to show that $$\sum_{cyc} a_i \leqslant \sum_{cyc} \frac{a_i^3}{a_{i+1}^2}$$ which follows easily from Rearrangement Inequality with the vectors $(a_i^3)$ and $\left(\dfrac1{a_i^2} \right)$ which are oppositely arranged.


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