work 014 – Korea 1998

Problem

For a,b,c positive reals such that a+b+c=abc ,

Prove that –

\sum \frac{1}{\sqrt{1+a^2}} \le \frac{3}{2}

Proof

The hypothesis implies that –

\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac} = 1

Set ,

x=\frac{1}{ab} , y=\frac{1}{bc} , z=\frac{1}{ac}

Thus, x+y+z=1

The inequality is now equivalent to –

\sum \sqrt{\frac{yz}{x+yz}} \le \frac{3}{2}

Since the sum of x,y,z is 1 –

\sum \sqrt{\frac{yz}{x+yz}} = \sum \sqrt{\frac{yz}{x(x+y+z)+yz}} = \sum \sqrt{\frac{yz}{(y+x)(y+z)}} = \sum \sqrt{\frac{y}{x+y}\cdot \frac{z}{y+z}}

Now by AM-GM , 

\sum \sqrt{\frac{y}{x+y}\cdot \frac{z}{y+z}} \le \frac{1}{2}\cdot [\sum \frac{y}{y+x}+\frac{z}{y+z}] = \frac{3}{2}

Equality occurs when x=y=z \implies a=b=c

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One comment

  1. incidentally…the same solution was posted in the ineq marathon by Dimitris X …. (ofcourse..i did not see his solution) … also … an almost similar solution is given in Hojoo Lee’s Pdf..

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