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 October 20th, 2019, 08:21 PM #1 Senior Member   Joined: Apr 2010 Posts: 456 Thanks: 1 inequality 2 Prove the following inequality ,where a,b,c are positive Nos: $\displaystyle \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \geq\frac{27}{2(a+b+c)^2}$ October 21st, 2019, 04:36 AM #2 Senior Member   Joined: Dec 2015 From: Earth Posts: 826 Thanks: 113 Math Focus: Elementary Math set $\displaystyle f(a,b,c)=\displaystyle \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} -\frac{27}{2(a+b+c)^2}\geq 0$. Since $\displaystyle f_a = f_b =f_c \;$ also $\displaystyle f_{a}'=f_{b}'=f_{c}'\geq 0$. set $\displaystyle a=b=c$, which holds for the bound of $\displaystyle f(a,b,c)$. $\displaystyle \frac{3}{2}\frac{1}{a^2 } - \frac{3}{2a^2 }=0\geq 0$. Last edited by idontknow; October 21st, 2019 at 05:08 AM. October 21st, 2019, 09:36 AM #3 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle It has a simple solution. No need for differentials... October 21st, 2019, 01:06 PM #4 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 Hint: it's easy to show that $\displaystyle (a + b + c)^2\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}\right) \geqslant 27$. Thanks from topsquark October 24th, 2019, 09:30 AM   #5
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 Originally Posted by skipjack Hint: it's easy to show that $\displaystyle (a + b + c)^2\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}\right) \geqslant 27$.
you mean that the initial inequality implies the above? October 24th, 2019, 02:38 PM #6 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 The AM-GM-HM inequality implies it, and I suspect that a slightly different use of that inequality can be used to prove the inequality you want proved. Thanks from idontknow October 26th, 2019, 05:32 AM #7 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 I've since confirmed my suspicion was correct. As the inequality isn't particularly sharp, an even shorter elementary solution may also exist, but could be time-consuming to find. October 26th, 2019, 08:04 AM   #8
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 Originally Posted by skipjack The AM-GM-HM inequality implies it .
How?

Last edited by skipjack; October 26th, 2019 at 10:44 AM. October 26th, 2019, 10:50 AM #9 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 $\displaystyle \left(\frac{(a + b + c)/3}{\sqrt[\Large3]{abc}}\right)^3 \geqslant 1 \implies (a + b + c)^2\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}\right) \geqslant 27$ Thanks from topsquark October 26th, 2019, 12:43 PM   #10
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 Originally Posted by skipjack $\displaystyle \left(\frac{(a + b + c)/3}{\sqrt[\Large3]{abc}}\right)^3 \geqslant 1 \implies (a + b + c)^2\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}\right) \geqslant 27$
Sorry, my mistake, I thοught the initial inequality was implying the above.

Last edited by skipjack; October 26th, 2019 at 06:21 PM. Tags inequality 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 StillAlive Calculus 5 September 3rd, 2016 12:45 AM sheran Algebra 6 July 28th, 2013 06:05 AM sachinrajsharma Algebra 3 February 16th, 2013 01:54 AM sachinrajsharma Algebra 1 February 15th, 2013 10:03 AM m.afifi Math Events 1 January 28th, 2013 08:58 AM

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