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 Algebra Pre-Algebra and Basic Algebra Math Forum

 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 $ab+bc+ca\leqslant a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$ $\implies$ $(a+b+c)^2\geqslant3(ab+bc+ca)$. Tags positive Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post cool012 Algebra 2 December 2nd, 2013 01:21 PM josellie1234 Number Theory 10 August 20th, 2013 09:28 AM wannabe1 Linear Algebra 1 April 26th, 2010 07:00 AM hello2413 Number Theory 3 March 15th, 2010 06:22 PM angelz429 Linear Algebra 1 May 7th, 2008 07:42 AM

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