1sry..my prev post was wrong....used calculus and got (a2/3+b2/3)3/2 same as aditya
but using a.m≥g.m i`m getting f(θ) ≥ (√2ab)(√sin3θ+cos3θ)....can anything be done after this?
\hspace{-16}$find minimum value of $f(\theta)=a\sec\theta+b\csc\theta$\\\\ $0 <\theta<\frac{\pi}{2}$
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262@ rishab - i didnt use AM≥GM. instead i used calculus and came to the result.
1708actually here a,b>0
aditiya and Debosmit you are right
(a^2/3 + b^2/3)^3/2
341By Holder's Inequality
(\sin^2 \theta + \cos^2 \theta) \left(\frac{b}{\sin \theta} + \frac{a}{\cos \theta} \right)\left(\frac{b}{\sin \theta} + \frac{a}{\cos \theta} \right) \ge \left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^3
and so we have
a \sec \theta + b \csc \theta \ge \left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}
