Let $f(x)=x+\frac 1x -\sqrt{2} ,\space x>0$ .
Then show that $f(f(x))-x$ is monotonically decreasing on $(\frac 1{\sqrt{2}} , 1)$
Seems a preety trivial question but have been trying for long without success.
1st Attempt : We have
$f'(x)=\big( 1-\frac 1{x^2} \big)<0 , \forall x \in (0,1) $
Now ,
$\frac d{dx}(f(f(x))-x)=\frac d{dx} \big( f(x) +\frac 1{f(x)} -\sqrt{2}-x \big) $
$= f'(x)\big( 1-\frac 1{f^2(x)}\big)-1 $
So I want $f^2(x) > 1$ which gives
$\big(x +\frac 1x \big)> \sqrt{2} +1 \space...(*)$
Taking equality in $(*)$ and solving the quadratic for $x$ , we get a rubbish expression for $x$
Attempt 2: I tried to simply show $\frac 1{\sqrt{2}} <x <y <1 \implies f(f(x))-x> f(f(y))-y $
But this was again unsuccessful.The graph can prove this result however it's apparently not so clear analytically .
Hints please.