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October 20th, 2019, 08:21 PM   #1
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inequality 2

Prove the following inequality ,where a,b,c are positive Nos:


$\displaystyle \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \geq\frac{27}{2(a+b+c)^2}$
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October 21st, 2019, 04:36 AM   #2
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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.
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October 21st, 2019, 09:36 AM   #3
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It has a simple solution. No need for differentials...
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October 21st, 2019, 01:06 PM   #4
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Hint: it's easy to show that $\displaystyle (a + b + c)^2\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}\right) \geqslant 27$.
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October 24th, 2019, 09:30 AM   #5
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Quote:
Originally Posted by skipjack View Post
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?
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October 24th, 2019, 02:38 PM   #6
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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.
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October 26th, 2019, 05:32 AM   #7
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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.
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October 26th, 2019, 08:04 AM   #8
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Quote:
Originally Posted by skipjack View Post
The AM-GM-HM inequality implies it .
How?

Last edited by skipjack; October 26th, 2019 at 10:44 AM.
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October 26th, 2019, 10:50 AM   #9
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$\displaystyle \left(\frac{(a + b + c)/3}{\sqrt[\Large3]{abc}}\right)^3 \geqslant 1 \implies (a + b + c)^2\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}\right) \geqslant 27$
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October 26th, 2019, 12:43 PM   #10
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Quote:
Originally Posted by skipjack View Post
$\displaystyle \left(\frac{(a + b + c)/3}{\sqrt[\Large3]{abc}}\right)^3 \geqslant 1 \implies (a + b + c)^2\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}\right) \geqslant 27$
Sorry, my mistake, I th╬┐ught the initial inequality was implying the above.

Last edited by skipjack; October 26th, 2019 at 06:21 PM.
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