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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.

Note that $$u=frac{b+1}{sqrt{1-b^2}+b-2}$$ so $$frac{du}{db}=frac{1big(sqrt{1-b^2}+b-2big)-(b+1)left(-frac{2b}{sqrt{1-b^2}}+1right)}{big(sqrt{1-b^2}+b-2big)^2}$$ and setting to zero gives $$-3sqrt{1-b^2}+b+1=0implies 1-b^2=frac{b^2+2b+1}9implies 5b^2+b-4=0$$ 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 $$bigg|-frac35+frac45-2bigg|>bigg|frac35+frac45-2bigg|$$

Try with $b=cos 2x$ and $a= sin 2x$.

begin{eqnarray}{b+1over a+b-2}&=& {2cos^2 xover -cos^2x+2sin x cos x -3sin^2x}\
&=& {2over -1+2tan x -3tan^2x}\
&=& {2over -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 $$ u= {2over -{2over 3}} = -3implies ....$$

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

$displaystyle bigg[u^2+(u-1)^2bigg]cdot bigg[a^2+b^2bigg]geq bigg[ua+(u-1)bbigg]^2$

So $displaystyle 2u^2-2u+1geq (1+2u)^2Rightarrow 2u^2+6uleq 0$

So $displaystyle 2u(u+3)leq 0Rightarrow -3 leq uleq 0$

So $displaystyle u^2 geq 9.$

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

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.