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$a,b,c$ 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.

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    $\begingroup$ What about AM-GM? $\endgroup$ – alex Mar 7 '14 at 10:55
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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 $a^2+b^2+c^2 \geq \sqrt{3}{abc}$.

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  • $\begingroup$ Can you elaborate the solution? I did not understand what you meant. $\endgroup$ – Hawk Mar 7 '14 at 11:44
  • $\begingroup$ 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. $\endgroup$ – user133281 Mar 7 '14 at 11:45
  • $\begingroup$ It does not need so much, it is immediate effect of Cauchy-Schwarz. I do not understand anything after 'AM-GM. Take the...' $\endgroup$ – Hawk Mar 7 '14 at 11:46
  • $\begingroup$ 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. $\endgroup$ – user133281 Mar 7 '14 at 11:48
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    $\begingroup$ @user133281 +1 Neat and Tidy :) $\endgroup$ – r9m Mar 7 '14 at 11:52
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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.

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