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April 7th, 2014, 10:51 AM   #1
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positive

a and b and c are positive number
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(a^2+2)(b^2+2)(c^2+2)/(ab +bc+ac)>=9
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April 7th, 2014, 03:37 PM   #2
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As stated it is incorrect. For very small a,b,c, the expression will be very small.

ab + bc + ac can be very small, while the other three terms together will be slightly > 8.
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April 7th, 2014, 03:53 PM   #3
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If ab + bc + ca is very small, then the expression on the left is very large!
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April 7th, 2014, 04:07 PM   #4
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$\displaystyle (a^2+2)(b^2+2)\,=\,(a^2+1)(b^2+1)+a^2+b^2+3$.

By Cauchy–Schwarz $(a^2+1)(b^2+1)\geqslant(a+b)^2$ and $2(a^2+b^2)=(1^2+1^2)(a^2+b^2)\geqslant(a+b)^2$.

$\displaystyle \therefore\ (a^2+2)(b^2+2)\,\geqslant\,(a+b)^2+\frac{(a+b)^2}2 +3\,=\,\frac32\left[(a+b)^2+2\right]$.

Hence

$$(a^2+2)(b^2+2)(c^2+2)$$
$\displaystyle \geqslant\ \frac32\left[(a+b)^2+2\right](2+c^2)$

$\displaystyle \geqslant\ \frac32\left[\sqrt2(a+b)+c\sqrt2\right]^2$ by Cauchy–Schwarz

$\displaystyle =\ 3(a+b+c)^2$

$\displaystyle \geqslant\ 9(ab+bc+ca)$ because by Cauchy–Schwarz $ab+bc+ca\leqslant a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$ $\implies$ $(a+b+c)^2\geqslant3(ab+bc+ca)$.
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