Alisha Pitts

2022-01-28

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

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$b=\frac{2-k}{2(k+1)}$ ... then again I got an equation in k which didnt

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

porekalahr

Beginner2022-01-29Added 10 answers

Try with $b=\mathrm{cos}2x\text{}\text{and}\text{}a=\mathrm{sin}2x$

$\frac{b+1}{a+b-2}=\frac{2{\mathrm{cos}}^{2}x}{-{\mathrm{cos}}^{2}x+2\mathrm{sin}x\mathrm{cos}x-3{\mathrm{sin}}^{2}x}$

$=\frac{2}{-1+2\mathrm{tan}x-3{\mathrm{tan}}^{2}x}$

$=\frac{2}{-1+2t-3{t}^{2}}$

where$t=\mathrm{tan}x$ . So the expression will take a minimum when quadratic function $g\left(t\right)=-3{t}^{2}+2t-1$ will take a maximum. Note that $g\left(t\right)<0\text{}\text{for all}\text{}t\in \mathbb{R}$

So

$u=\frac{2}{-\frac{2}{3}}=-3\Rightarrow \dots .$

where

So

Palandriy0

Beginner2022-01-30Added 14 answers

Note that

$u=\frac{b+1}{\sqrt{1-{b}^{2}}+b-2}$

so

$\frac{du}{db}=\frac{1(\sqrt{1-{b}^{2}}+b-2)-(b+1)(-\frac{2b}{\sqrt{1-{b}^{2}}}+1)}{{(\sqrt{1-{b}^{2}}+b-2)}^{2}}$

and setting to zero gives

$-3\sqrt{1-{b}^{2}}+b+1=0\Rightarrow 1-{b}^{2}=\frac{{b}^{2}+2b+1}{9}\Rightarrow 5{b}^{2}+b-4=0$

and we see that$b=\frac{4}{5},-1$ are roots.

Checking second derivatives, we have that 4/5 is a minimum.

Hence

${u}^{2}={\left(\frac{\frac{45}{+}1}{\frac{35}{+}\frac{45}{-}2}\right)}^{2}=9$

Note that the negative root (-3/5) is also possible, but that yields a lower value of$u}^{2$

$|-\frac{3}{5}+\frac{4}{5}-2|>|\frac{3}{5}+\frac{4}{5}-2|$

so

and setting to zero gives

and we see that

Checking second derivatives, we have that 4/5 is a minimum.

Hence

Note that the negative root (-3/5) is also possible, but that yields a lower value of