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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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Quote:
 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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Quote:
 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.

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