2021/006) Let a,b,c be real with $a,b,c > 1$ and $\frac{1}{a^2-1} + \frac{1}{b^2-1} + \frac{1}{c^2-1} = 1$ Show that $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} <= 1$

We have $\frac{1}{a^2-1} + \frac{1}{b^2-1} + \frac{1}{c^2-1} = 1\cdots(1)$

Now 

$\frac{2}{a^2-1} = \frac{1}{a+1} + \frac{1}{a-1}$

As $a > 1$ so we have $a-1 > 0$ and hence

$\frac{1}{a+1} <  \frac{1}{a-1}$

adding  $\frac{1}{a+1}$ on both sides

$\frac{2}{a+1} <  \frac{1}{a-1} + \frac{1}{a+1}$

Or $\frac{2}{a+1} >  \frac{2}{a^2-1}$

Or $\frac{1}{a+1} <  \frac{1}{a^2-1}\cdots(2)$

Similarly

Or $\frac{1}{b+1} <  \frac{1}{a^2-1}\cdots(3)$

Or $\frac{1}{c+1} <  \frac{1}{a^2-1}\cdots(4)$ 

Adding above 3 we get

$\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} <  \frac{1}{a^2-1} + \frac{1}{b^2-1} + \frac{1}{c^2-1}$

or $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} <  1$