# $a,b,c$ are positive real numbers such that, $a+b+c\ge abc$. Prove that $a^2+b^2+c^2\ge \sqrt{3}abc$

$a,b,c$ are positive real numbers such that, $a+b+c\ge abc$. Prove that $a^2+b^2+c^2\ge \sqrt{3}abc$

My work:
I tried using Cauchy-Schwarz inequality to find that,
$(a^2+b^2+c^2)(1^2+1^2+1^2)\ge (a+b+c)^2$
$(a^2+b^2+c^2)\ge \dfrac13(a+b+c)^2$
$\sqrt{(a^2+b^2+c^2)}\ge \dfrac{1}{\sqrt{3}}(a+b+c)\ge \dfrac{1}{\sqrt{3}}abc$
which is not what I need and neither I can use it to prove the required inequality. Please help.

• What about AM-GM? – alex Mar 7 '14 at 10:55

We have $a^2 + b^2 + c^2 \geq \frac{(a+b+c)^2}{3} \geq \frac{(abc)^2}{3}$ and $a^2+b^2+c^2 \geq 3 \sqrt[3]{a^2b^2c^2}$ by AM-GM. Take the $1/4$-th power of the first inequality and the $3/4$-th power of the second inequality, and multiply (this is allowed since everything is positive). The result is $a^2+b^2+c^2 \geq \sqrt{3}{abc}$.

• Can you elaborate the solution? I did not understand what you meant. – Hawk Mar 7 '14 at 11:44
• Please point out the steps you do not understand. For the first inequality, we use $3(a^2+b^2+c^2) \geq (a+b+c)^2$, which follows from $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$. For the second inequality, we use AM-GM directly. – user133281 Mar 7 '14 at 11:45
• It does not need so much, it is immediate effect of Cauchy-Schwarz. I do not understand anything after 'AM-GM. Take the...' – Hawk Mar 7 '14 at 11:46
• We find $(a^2+b^2+c^2)^{1/4} \geq \frac{\sqrt{abc}}{(3^{1/4})}$ and $(a^2+b^2+c^2)^{3/4} \geq 3^{3/4} \sqrt{abc}$. Multiplying these inequalities gives the desired result. – user133281 Mar 7 '14 at 11:48
• @user133281 +1 Neat and Tidy :) – r9m Mar 7 '14 at 11:52

here is another way

$$(a^2+b^2+c^2)^2=\sum\limits_{cyc} a^4 + \sum\limits_{cyc} 2a^2b^2 \ge abc(a+b+c) + 2abc(a+b+c) =3abc(a+b+c)\ge3a^2b^2c^2$$

inequality $\sum\limits_{cyc} a^4 \ge \sum\limits_{cyc} a^2b^2 \ge abc(a+b+c)$ can be proved using AM-GM or Cauchy-Schwarz.