Minimum value of b+1a+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=2−k2(k+1)... then again I got an equation in k which didnt

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Minimum value of b+1a+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=2−k2(k+1)... then again I got an equation in k which didnt

porekalahr · Accepted Answer

Try with b=cos⁡2x  and  a=sin⁡2x  b+1a+b−2=2cos2x−cos2x+2sin⁡xcos⁡x−3sin2x  =2−1+2tan⁡x−3tan2x  =2−1+2t−3t2  where t=tan⁡x. So the expression will take a minimum when quadratic function g(t)=−3t2+2t−1 will take a maximum. Note that g(t)&amp;lt;0  for all  t∈R  So  u=2−23=−3⇒….

Palandriy0 · Accepted Answer

Note that  u=b+11−b2+b−2  so  dudb=1(1−b2+b−2)−(b+1)(−2b1−b2+1)(1−b2+b−2)2  and setting to zero gives  −31−b2+b+1=0⇒1−b2=b2+2b+19⇒5b2+b−4=0  and we see that b=45,−1 are roots.   Checking second derivatives, we have that 4/5 is a minimum.   Hence  u2=(45+135+45−2)2=9  Note that the negative root (-3/5) is also possible, but that yields a lower value of u2  |−35+45−2|&amp;gt;|35+45−2|

Minimum value of \frac{b+1}{a+b-2} Attempt: Then I tried this way: Let a=bk

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