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 April 7th, 2014, 10:51 AM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 positive a and b and c are positive number prooof (a^2+2)(b^2+2)(c^2+2)/(ab +bc+ac)>=9
 April 7th, 2014, 03:37 PM #2 Global Moderator   Joined: May 2007 Posts: 6,834 Thanks: 733 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.
 April 7th, 2014, 03:53 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,689 Thanks: 2669 Math Focus: Mainly analysis and algebra If ab + bc + ca is very small, then the expression on the left is very large!
 April 7th, 2014, 04:07 PM #4 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra $\displaystyle (a^2+2)(b^2+2)\,=\,(a^2+1)(b^2+1)+a^2+b^2+3$. 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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