prove that $\sum_{cyc}\frac{a^2}{b}\ge 4$

Question

prove that $$\sum_{cyc}\frac{a^2}{b}\ge 4$$ if $a,b,c,d\ge 0$,$a^2+b^2+c^2+d^2=4$

My try:

by Titu's lemma

$$\sum_{cyc}\frac{a^2}{b}\ge a+b+c+d$$

now from given condition we can say $a\ge 0,a\le 2$,,or $a(a-2)\le 0$ or $a^2\le2a$

by this i was only able to prove that

$$\sum_{cyc}\frac{a^2}{b}\ge 2$$

score 1 · Accepted Answer · answered Sep 02 '20 at 15:51

1

By Holder $$\left(\sum_{cyc}\frac{a^2}{b}\right)^2\sum_{cyc}a^2b^2\geq\left(\sum_{cyc}a^2\right)^3.$$ Can you end it now?

The hint: use AM-GM.

answered Sep 02 '20 at 15:51

Michael Rozenberg

thank you (+1),,yes i get it , but why isn't my method also giving the answer – Albus Dumbledore Sep 02 '20 at 15:59
@Quantum Because $a+b+c+d\leq4$. It's not enough. – Michael Rozenberg Sep 02 '20 at 16:14

1 Answers1