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.

  • 5
    $\begingroup$ It's correct, except that you should replace the $\implies$ by $\iff$. $\endgroup$ – 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 $A\implies B \implies C$).

This is one of the reasons why it is always better to use words rather than mathematical notations when it comes to reasonning, in order to avoid confusion, and simply because it (hopefully) requires less effort from the ones who read you since it makes your proof much smoother.

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Alternatively by $AM \ge GM$,

\begin{align*} a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2) &= a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2 \\ & \ge 6\sqrt[6]{a^6b^6c^6} \\ \end{align*}

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