IMO LONGLISTED PROBLEM 1970
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$\frac{bc}{b+c} $+$\frac{ca}{c+a}$+$\frac{ab}{a+b}$ $\ge$ $\frac{a+b+c}{2}$.where a,b,c>0.
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- Posts:190
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Re: IMO LONGLISTED PROBLEM 1970
a very nice problem.
Re: IMO LONGLISTED PROBLEM 1970
Let, $a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}$
Then the inequality is reduced to \[\sum \frac{1}{x+y}\ge \frac{xy+yz+zx}{xyz}=\sum \frac{1}{2z}\]
This is true by re-arrangement inequality.
Then the inequality is reduced to \[\sum \frac{1}{x+y}\ge \frac{xy+yz+zx}{xyz}=\sum \frac{1}{2z}\]
This is true by re-arrangement inequality.
One one thing is neutral in the universe, that is $0$.
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- Posts:190
- Joined:Sat Apr 23, 2011 8:55 am
- Location:Khulna
Re: IMO LONGLISTED PROBLEM 1970
thanks 4 a very nice,small solution.my solution is too long.