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.
 A: Note that $$u=\frac{b+1}{\sqrt{1-b^2}+b-2}$$ so $$\frac{du}{db}=\frac{1\big(\sqrt{1-b^2}+b-2\big)-(b+1)\left(-\frac{2b}{\sqrt{1-b^2}}+1\right)}{\big(\sqrt{1-b^2}+b-2\big)^2}$$ and setting to zero gives $$-3\sqrt{1-b^2}+b+1=0\implies 1-b^2=\frac{b^2+2b+1}9\implies 5b^2+b-4=0$$ and we see that $b=4/5,-1$ are roots.
Checking second derivatives, we have that $4/5$ is a minimum.
Hence $$u^2=\left(\frac{\frac45+1}{\frac35+\frac45-2}\right)^2=9.$$

Note that the negative root ($-3/5$) is also possible, but that yields a lower value of $u^2$ since $$\bigg|-\frac35+\frac45-2\bigg|>\bigg|\frac35+\frac45-2\bigg|$$
A: Let $\displaystyle u=\frac{b+1}{a+b-2}\Rightarrow ua+ub-2u=b+1$
So $ua+(u-1)b=1+2u$
Now Using Cauchy Schwarz Inequality
$\displaystyle \bigg[u^2+(u-1)^2\bigg]\cdot  \bigg[a^2+b^2\bigg]\geq \bigg[ua+(u-1)b\bigg]^2$
So $\displaystyle 2u^2-2u+1\geq (1+2u)^2\Rightarrow 2u^2+6u\leq 0$
So $\displaystyle 2u(u+3)\leq 0\Rightarrow -3 \leq u\leq 0$
So $\displaystyle u^2 \geq 9.$
A: A bit geometry;
1)$x^2+y^2 = 1$, a circle, centre $(0,0)$, $r=1$.
2) Minimum of $C$:
$C:=\dfrac{y+1}{x+y-2}$
(Note: $x+y-2 \not =0$).
$C(x+y-2) = y+1$, or
$Cx +y(C-1) -(2C+1)= 0$, a straight line.
The line touches or intersects the circle 1)
if the distance line-to-origin $\le 1$ (radius).
Distance to $(0,0):$
$d =\dfrac{|2C+1|}{\sqrt{C^2+(C-1)^2}} \le 1.$
$(2C+1)^2 \le C^2 + (C-1)^2;$
$4C^2 +4C +1 \le 2C^2 -2C+1;$
$2C(C+3) \le 0.$
Hence: $-3 \le C \le 0$.
Minimum at $C =-3$.
Used: Line to point distance formula:
http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html
A: Just for a variation, using Lagrange’s method:
$$
f(a,b,t)=\frac{b+1}{a+b-2}-t(a^2+b^2-1)
$$
Then
\begin{align}
\frac{\partial f}{\partial a}&=-\frac{b+1}{(a+b-2)^2}-2at \\[6px]
\frac{\partial f}{\partial b}&=\frac{a-3}{(a+b-2)^2}-2bt
\end{align}
If these equal $0$, then
$$
-\frac{b+1}{a(a+b-2)^2}=\frac{a-3}{b(a+b-2)^2}
$$
so that $-b^2-b=a^2-3a$, that gives $3a-b=1$ and from $a^2+b^2=1$ we derive $a=0$ or $a=3/5$.
The critical points are thus $(0,-1)$ and $(3/5,4/5)$. We have
$$
f(0,-1,0)=\frac{2}{3},\quad
f(3/5,4/5,0)=-3,\quad
$$
This also shows the maximum.
A: 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 $9$.
A: From
$$
\frac{b+1}{a+b-2}= u\Rightarrow L\to a = 2-b\frac{b+1}{u}
$$
So $L$ should be tangent to $a^2+b^2=1\;$ then substituting we have the condition
$$
(2b^2-4b+3)u^2+(4+2b-2b^2)u+(b+1)^2 = 0
$$
and solving for $u$
$$
u = \frac{(b+1)^2}{b^2-b\pm\sqrt{(1-b) (b+1)^3}-2}
$$
but tangency implies on $\sqrt{(1-b) (b+1)^3}=0\;$ hence the solutions for tangency are $b = \pm 1$ etc.
A: Try with $b=\cos 2x$ and $a= \sin 2x$.
\begin{eqnarray}{b+1\over a+b-2}&=& {2\cos^2 x\over  -\cos^2x+2\sin x \cos x -3\sin^2x}\\
&=& {2\over  -1+2\tan x  -3\tan^2x}\\
&=& {2\over  -1+2t  -3t^2}
\end{eqnarray}
where $t= \tan x $. So the expression will take a minimum when quadratic function $g(t)=-3t^2+2t-1$ will take a maximum. Note that $g(t)<0$ for all $t \in \mathbb{R}$ 
So $$ u= {2\over -{2\over 3}} = -3\implies ....$$
