min value

sry..my prev post was wrong....used calculus and got (a^2/3+b^2/3)^3/2 same as aditya

but using a.m≥g.m i`m getting f(θ) ≥ (√2ab)(√sin³Î¸+cos³Î¸)....can anything be done after this?

i got (a^2/3 + b^2/3)^3/2

@aditya its not given that a,b>0 so i guess you can't use a.m.>g.m.

@ rishab - i didnt use AM≥GM. instead i used calculus and came to the result.

@rishabh:my mistake....can`t use am≥gm....

oops sorry. i meant debosmit, not aditya.

actually here a,b>0

aditiya and Debosmit you are right

(a^2/3 + b^2/3)^3/2

By 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}}

-b

Thanks bhatt sir,aditiya,Debosmit