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

Another way.
We need to prove that
$$(a+b+c)^2\geq2\sum_{cyc}a^2b+abc(a+b+c)$$ and since $$\frac{3}{a^2+b^2+c^2}\geq1$$ and $$1\geq\frac{a^2+b^2+c^2}{3},$$ it's enough to prove that:
$$(a+b+c)^2\sqrt{\frac{a^2+b^2+c^2}{3}}\geq2\sum_{cyc}a^2b+\sqrt{\frac{3}{a^2+b^2+c^2}}abc(a+b+c)$$ or
$$(a+b+c)^2(a^2+b^2+c^2)\geq2\sum_{cyc}a^2b\sqrt{3(a^2+b^2+c^2)}+3abc(a+b+c).$$ 
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v>0$, $abc=w^3$ and $u=tv$.
Thus, $t\geq1$ and since by Rearrangement $\sum_{cyc}a^2b\leq4u^3-w^3$, it's enough to prove that:
$$9u^2(9u^2-6v^2)\geq2(4u^3-w^3)\sqrt{3(9u^2-6v^2)}+9uw^3$$ or
$$9u^2(3u^2-2v^2)-8u^3\sqrt{3u^2-2v^2}\geq(3u-2\sqrt{3u^2-2v^2})w^3.$$
Now, if $$(3u-2\sqrt{2u^2-2v^2}<0$$ so our inequality is obviously true because
$$9u^2(3u^2-2v^2)-8u^3\sqrt{3u^2-2v^2}=$$
$$=u^2\sqrt{3u^2-2v^2}(9\sqrt{3u^2-2v^2}-8u)\geq u^2\sqrt{3u^2-2v^2}(9\sqrt{u^2}-8u)>0.$$
But for $3u-2\sqrt{2u^2-2v^2}\geq0$ since $w^3\leq v^3,$ it's enough to prove that
$$9t^2(3t^2-2)-8t^3\sqrt{3t^2-2}\geq3t-2\sqrt{3t^2-2}$$ or
$$27t^4-18t^2-3t\geq(8t^3-2)\sqrt{3t^2-2}$$ or
$$(t-1)(537t^7+537t^6-307t^5-373t^4-49t^3-5t^2-8t-8)\geq0,$$ which is obvious for $t\geq1$.
