# Prove $a^{2}(1+b^{2})+b^{2}(1+c^{2})+c^{2}(1+a^2)\geq 6abc$

Prove $a^{2}(1+b^{2})+b^{2}(1+c^{2})+c^{2}(1+a^2)\geq 6abc$

My attempt:

$a^{2}(1+b^{2})+b^{2}(1+c^{2})+c^{2}(1+a^2)-6abc\geq 0$

$\implies a^{2}+a^{2}b^{2}+b^{2}+b^{2}c^{2}+c^{2}+c^{2}a^{2}-2abc-2abc-2abc\geq 0$

$\implies (a-bc)^{2}+(b-ac)^{2}+(c-ab)^{2}\geq 0$

Each of these terms must be non-negative, thus the sum is also non-negative.

I'm new to writing proofs, so I don't know whether this proof is fine.

• It's correct, except that you should replace the $\implies$ by $\iff$. – GoodDeeds Oct 30 '16 at 10:09

The idea behind your proof is okay, except for the fact that your $\implies$ should rather be $\iff$ for it to be sufficient.
What's more it is not (formally) correct to write $\implies$ signs one after the other, since $\implies$ is not associative ($(A\implies B) \implies C$ is different from $A\implies (B \implies C)$, and both are different from what you seem to mean by $A\implies B \implies C$).
Alternatively by $AM \ge GM$,