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 $\displaystyle (a + b + c)^2\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}\right) \geqslant 27$.

October 24th, 2019, 09:30 AM  #5 
Senior Member Joined: Apr 2010 Posts: 456 Thanks: 1  
October 24th, 2019, 02:38 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 21,110 Thanks: 2326 
The AMGMHM inequality implies it, and I suspect that a slightly different use of that inequality can be used to prove the inequality you want proved.

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 timeconsuming to find. 
October 26th, 2019, 08:04 AM  #8 
Senior Member Joined: Apr 2010 Posts: 456 Thanks: 1  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 
$\displaystyle \left(\frac{(a + b + c)/3}{\sqrt[\Large3]{abc}}\right)^3 \geqslant 1 \implies (a + b + c)^2\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}\right) \geqslant 27$

October 26th, 2019, 12:43 PM  #10 
Senior Member Joined: Apr 2010 Posts: 456 Thanks: 1  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  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Triangle Inequality: Prove Absolute Value Inequality  StillAlive  Calculus  5  September 3rd, 2016 12:45 AM 
inequality  sheran  Algebra  6  July 28th, 2013 06:05 AM 
Inequality  sachinrajsharma  Algebra  3  February 16th, 2013 01:54 AM 
Inequality ii  sachinrajsharma  Algebra  1  February 15th, 2013 10:03 AM 
inequality  m.afifi  Math Events  1  January 28th, 2013 08:58 AM 