Minimum value of $\frac{b+1}{a+b-2}$ 
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $\dfrac{b+1}{a+b-2}$, then find the value of $u^2$. 

Attempt: 
Then I tried this way: Let $a= bk$ for some real $k$. 
Then I got $f(b)$  in terms of b and k which is minmum when $b = \dfrac{2-k}{2(k+1)}$ ... then again I got an equation in $k$ which didn't simplify. 
Please suggest an efficient  way to solve it.

Pulling a small rabbit from a hat, consider 
$$f(a,b)=3+{b+1\over a+b-2}={3a+4b-5\over a+b-2}$$
It's clear that $a^2+b^2=1$ implies $a+b-2\lt0$. By Cauchy-Schwartz, we have
$$(3a+4b)^2\le(a^2+b^2)(3^2+4^2)=25=5^2$$
and therefore $3a+4b-5\le0$ if $a^2+b^2=1$. Thus $f(a,b)\ge0$ for all $a$ and $b$ for which $a^2+b^2=1$. But also $f({3\over5},{4\over5})=0$, so the minimum value of $(b+1)/(a+b-2)=f(a,b)-3$ with $a^2+b^2=1$ is $-3$, the square of which is $9$.
