# Prove the following inequality using am gm inequality

Problem:

If $$abc \le a+b+c$$ then prove that $$a^2+b^2+c^2\ge √3abc$$

I squared the inequality to get $$a^2+b^2+c^2+2ab+2bc+2ac\ge a^2b^2c^2$$. Now I tried to apply am gm but it didn't work.I have to someway end up with $$3a^2b^2c^2$$ on rhs and then it will be easy.Any help

If $$abc\leq0$$, so it's obvuios.

Let $$abc>0$$.

Thus, $$\frac{a+b+c}{abc}\geq1$$ and we need to prove that: $$(a^2+b^2+c^2)^2\geq3a^2b^2c^2,$$ for which it's enough to prove that: $$(a^2+b^2+c^2)^2\geq3a^2b^2c^2\cdot\frac{a+b+c}{abc}$$ or $$\sum_{cyc}(a^4+2a^2b^2-3a^2bc)\geq0$$ or $$\sum_{cyc}(2a^4-2a^2b^2+6a^2b^2-6a^2bc)\geq0$$ or $$\sum_{cyc}(a^4-2a^2b^2+b^4+3a^2c^2-6c^2ab+3b^2c^2)\geq0$$ or $$\sum_{cyc}(a-b)^2((a+b)^2+3c^2)\geq0$$ and we are done!

For positive variables a proof a bit of shorter.

By AM-GM we obtain: $$a^2+b^2+c^2=\sqrt{(a^2+b^2+c^2)^2}\geq\sqrt{3(a^2b^2+a^2c^2+b^2c^2)}=$$ $$=\sqrt{\frac{3}{2}\sum_{cyc}(a^2b^2+a^2c^2)}\geq \sqrt{\frac{3}{2}\sum_{cyc}2a^2bc}=\sqrt{3abc(a+b+c)}\geq\sqrt3abc.$$

• Also,can we use am gm somewhere to skip some steps?? – Mathematical Curiosity Mar 17 '20 at 17:44
• @Mathematical Curiosity In the end of the proof we can use AM-GM because it's enough to assume that $a$, $b$ and $c$ are positives. But we can not say it before because $a$, $b$ and $c$ can be negatives and we can not use AM-GM in this case. – Michael Rozenberg Mar 17 '20 at 18:44
• Then let us assume that a,b,c are positive then what other way can we take?? – Mathematical Curiosity Mar 17 '20 at 19:46
• @Mathematical Curiosity Yes, of course. But it would be a bit of easier inequality. – Michael Rozenberg Mar 17 '20 at 21:57
• Rosenberg can you show it briefly?? – Mathematical Curiosity Mar 18 '20 at 5:26