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

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

Answer 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.