<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] Assume w.l.o.g that $a \leq b \leq c$.
Then we have : $(a+b+c)^3 \geq 7(a^2b+b^2c+c^2a)+6abc$; thus it is sufficient to prove: $(a+b+c)(a+b+c-abc) \geq \dfrac{2}{7}((a+b+c)^3-6abc)$.
Let $x=a+b+c$ and $y=abc$. The last inequality is then: $x^2-xy \geq \dfrac{2}{7}(x^3-6y)$, which is equivalent to: $x^2+\dfrac{12y}{7} \geq \dfrac{2x}{7}(x^2+\dfrac{7y}{2})$ and since $x \leq 3$ (using Cauchy-Schwarz on the original inequality), it is sufficient to prove: $x^2+\dfrac{12y}{7} \geq \dfrac{6}{7}(x^2+\dfrac{7y}{2})$, which is equivalent to: $x^2 \geq 9y$, which follows from A.M.-G.M. ($x \geq 3y^{1/3} \geq 3y^{1/2}$) where we used $y=abc\leq 1$.
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