2

Let $a, b, c, d > 0 \in \mathbb{R}$ such that $a^2 + b^2 + c^2 + d^2 = 4$. Show that:

$S = \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq 4$

My approach: I used the Cauchy-Schwarz inequality to show that $S \geq a + b + c + d$ but that is useless as $a + b + c + d \leq 4$. How would you approach this problem? (Only hints desired)

iart
  • 383
  • 1
  • 7

1 Answers1

3

By Holder's Inequality, $$S^2\sum_{cyc} a^2b^2\ge (a^2+b^2+c^2+d^2)^3$$

So it remains to note that by AM-GM $$\sum_{cyc}a^2b^2 = (a^2+c^2)(b^2+d^2) \le \frac{\left((a^2+c^2)+(b^2+d^2)\right)^2}4$$

Macavity
  • 46,381