Prove that $a^2+b^2+c^2 \geq 3 abc$ given that $a+b+c \geq 3 abc$
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2Again, nature of $a,b,c?$ – lab bhattacharjee Nov 27 '13 at 10:40
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Next time, please show us what you have tried. :) – Ahaan S. Rungta Nov 27 '13 at 14:13
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If $a,b,c\in \Bbb R$, and if $a+b+c\ge 3abc$, then: $$a^2+b^2+c^2\ge 3abc$$
Proof:
Case 1:
If $abc\le 1$, then we use AM-GM inequality, and have: $$a^2+b^2+c^2=|a|^2+|b|^2+|c|^2\ge 3\sqrt[3]{|abc|^2}\ge 3abc$$ this is true because $$\Longleftrightarrow |abc|^2\ge (abc)^3$$
Case 2:
If $abc\ge 1$, since: $$a+b+c\ge 3abc>0$$ we can use Cauchy-Schwarz inequality in the following way: $$a^2+b^2+c^2\ge\dfrac{1}{3}(a+b+c)^2\ge 3a^2b^2c^2=3(|abc|)^2\ge 3abc$$
