How to prove the following using Cauchy Schwarz inequality.

Question

If $a$, $b$ and $c$ are non-negative numbers, then prove the following by using Cauchy Schwarz inequality. $$\sqrt{3a^2+ab}+ \sqrt{3b^2+bc}+ \sqrt{3c^2+ca} \leq 2(a+b+c)$$ How to think for this?

Answer 1

Hint: $\sqrt{3a^2+ab} = \left(\sqrt{a}\right)\left(\sqrt{3a+b}\right)$. Can you figure out how to apply Cauchy-Schwarz from here?

Answer 2

For non-negative variables by C-S $$\sum_{cyc}\sqrt{3a^2+ab}=\sum_{cyc}\sqrt{a(3a+b)}\leq\sqrt{\sum_{cyc}a\sum_{cyc}(3a+b)}=2(a+b+c).$$

Also, AM-GM works: $$\sum_{cyc}\sqrt{3a^2+ab}=\frac{1}{4}\sum_{cyc}\left(2\sqrt{4a(3a+b)}\right)\leq\frac{1}{4}\sum_{cyc}(4a+3a+b)=2(a+b+c).$$