Remark. I really like inequality problems, but I've found that I have severe difficulties with ones that are very elegant form. I can prove something simple like this $$\color{black}{\sqrt{3a+bc}+\sqrt{3b+ca} +\sqrt{3c+ab}\le 6; a,b,c\ge 0: a^2+b^2+c^2=3. }$$
It is obvious true by Cauchy-Schwarz $$\sum_{cyc}\sqrt{3a+bc}\le \sqrt{3(3a+3b+3c+ab+bc+ca)}\le 6$$ since $a+b+c\le \sqrt{3(a^2+b^2+c^2)}=3$ and $ab+bc+ca\le a^2+b^2+c^2=3.$
The following is a bit of harder.
If $a,b,c\ge 0: a^2+b^2+c^2=3.$ Prove that $$\color{black}{\sqrt{3a^3+bc}+\sqrt{3b^3+ca} +\sqrt{3c^3+ab}\le 6. }$$
I saw that problem in AOPS. Unfortunately, it was deleted by proposer.
In the past, I often homogenize before try other approach. For this case, it would not be nice.
I tried to use Cauchy-Schwarz inequality without success.
Indeed, $$\sum_{cyc}\sqrt{3a^3+bc}\le \sqrt{3(3a^3+3b^3+3c^3+ab+bc+ca)}\le 6$$is already wrong when $a=b=0; c=\sqrt{3}.$
Now, we need to find a better C-S using. For example, $$\sum_{cyc}\sqrt{3a^3+bc}=\sum_{cyc}\sqrt{\frac{3a^3+bc}{b+c}\cdot (b+c)}\le \sum_{cyc}\sqrt{\frac{3a^3+bc}{b+c}\cdot 2(a+b+c)}.$$ Thus, it's enough to prove $$\frac{3a^3+bc}{b+c}+\frac{3b^3+ca}{a+c}+\frac{3c^3+ab}{b+a}\le \frac{18}{a+b+c}.$$
I was stuck here. Can you help me continue my idea? I'm very appreciate your help.
Also, could you share some approach to solve same kind of this inequality which is difficult to homogenize.
By C-S $$\sum_{cyc}\sqrt{3a^3+bc}\leq\sqrt{\sum_{cyc}\frac{3a^3+bc}{3a+2}\sum_{cyc}(3a+2)}$$ and it's enough to prove that: $$\sum_{cyc}\frac{3a^3+bc}{3a+2}\leq\frac{12}{a+b+c+2},$$ which is a linear inequality of $w^3$.
Thus, by $uvw$(see here: https://artofproblemsolving.com/community/c6h278791) it's enough to prove the last inequality in the following two cases.
$abc=w^3=0.$
Two variables are equal.
I got that in these cases the inequality is true.
Can you get a full proof now?