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

porekalahr

Try with
$\frac{b+1}{a+b-2}=\frac{2{\mathrm{cos}}^{2}x}{-{\mathrm{cos}}^{2}x+2\mathrm{sin}x\mathrm{cos}x-3{\mathrm{sin}}^{2}x}$
$=\frac{2}{-1+2\mathrm{tan}x-3{\mathrm{tan}}^{2}x}$
$=\frac{2}{-1+2t-3{t}^{2}}$
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
So
$u=\frac{2}{-\frac{2}{3}}=-3⇒\dots .$

Palandriy0

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

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