Let $a,b,c\ge 0: a^2+b^2+c^2=3.$ Prove that$$\color{black}{\sqrt{a+4b}+\sqrt{b+4c}+\sqrt{c+4a}\le 3\sqrt{abc+4}. }$$
Naturally, I use Cauchy-Schwarz as $$\sqrt{a+4b}+\sqrt{b+4c}+\sqrt{c+4a}\le \sqrt{15(a+b+c)},$$ but $$\color{black}{15(a+b+c)\le 9abc+36,}$$ is already wrong when $a=b=\sqrt{\dfrac{3}{2}}; c=0.$
Hope you help me give some ideas and how it motivates a proof.
By C-S $$\sum_{cyc}\sqrt{a+4b}\leq\sqrt{\sum_{cyc}\frac{a+4b}{3a+8b+2c}\sum_{cyc}(3a+8b+2c)}$$ and it's enough to prove that $$\sum_{cyc}\frac{a+4b}{3a+8b+2c}\leq\frac{9(abc+4)}{13(a+b+c)}.$$ The rest is smooth, but my solution is not so nice.
Can you end it now?