# 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 $x=y=z \implies a=b=c$