<s>[INST] How to prove this inequality? $a^{2}+b^{2}+c^{2}\leq 3$ How to prove this inequality?
If $a^{2}+b^{2}+c^{2}\leq 3$ and $a,b,c\in \Bbb R^+$, then
$$\left( a+b+c\right) \left( a+b+c-abc\right)\geq  2\left( a^{2}b+b^{2}c+c^{2}a\right) $$
I tried AM>GM but I couldn't get result
 [/INST] Alternative solution:
Recall the known inequality $a^2b+b^2c+c^2a \le \frac{4}{27}(a+b+c)^3 - abc$ for $a, b, c \ge 0$.
See: https://artofproblemsolving.com/community/c6h478168p2677448, 
https://artofproblemsolving.com/community/c4h1705936p10979968, 
https://artofproblemsolving.com/community/c6h479531p2685016
It suffices to prove that
$(a+b+c)(a+b+c-abc) \ge \frac{8}{27}(a+b+c)^3 - 2abc$
or
$$(a+b+c)^2 + 2abc \ge \frac{8}{27}(a+b+c)^3 + (a+b+c)abc.$$
By using the method of @Aravind,  since $a+b+c \le 3\sqrt{\frac{a^2+b^2+c^2}{3}} \le 3$, it suffices to prove that
$$(a+b+c)^2 + 2abc \ge \frac{8}{27}(a+b+c)^2 \cdot 3 + 3abc$$
or
$$(a+b+c)^2 \ge 9abc.$$
Since $\frac{a+b+c}{3} \ge \sqrt[3]{abc}$ and $a+b+c\le 3$, we have $abc\le 1$ and $(a+b+c)^2 \ge 9(abc)^{2/3} \ge 9abc$. 
We are done.
</s>