Minimum value of $\frac{b+1}{a+b-2}$

Question

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.

Answer 1

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 $$u^2=\left(\frac{\frac45+1}{\frac35+\frac45-2}\right)^2=9.$$

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Note that the negative root ($-3/5$) is also possible, but that yields a lower value of $u^2$ since $$\bigg|-\frac35+\frac45-2\bigg|>\bigg|\frac35+\frac45-2\bigg|$$

Answer 2

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 $$ u= {2\over -{2\over 3}} = -3\implies ....$$

Answer 3

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.

Answer 4

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 $\displaystyle u^2 \geq 9.$

Answer 5

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.

Answer 6

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

Answer 7

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 $9$.