MQ74 --- Inequality

Let f(θ) = acscθ + bsecθ

f'(θ) = bsecθtanθ - acscθcotθ

= b(1/cosθ)(sinθ/cosθ) - a(1/sinθ)(cosθ/sinθ)

= [b(sinθ)^3 - a(cosθ)^3]/[(sinθcosθ)^2]

when f'(θ) = 0, f(θ) has the min. value

f'(θ) = 0 => b(sinθ)^3 - a(cosθ)^3 = 0 => (tanθ)^3 = a/b

Then cscθ = √(a^(2/3) + b^(2/3))/a^(1/3), secθ = √(a^(2/3) + b^(2/3))/b^(1/3)

The minimum of acscθ + bsecθ in terms of a, b is

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

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

似曾相識, Myisland 復仇記~
http://hk.knowledge.yahoo.com/question/question?qid=7011051001326

Given a,b>0 and Q∈(0,π/2), find the minimum & maximum value of Let u=a^(1/3), v=b^(1/3), c=cosQ, s=sinQy(Q)=a*cscQ+b*secQ.Let 0=y'(Q)=-a*cscQ*cotQ+b*secQ*tanQ=-a*c/s^2+b*s/c^2=> a*c^3=b*s^30=(u*c)^3-(v*s)^3=(u*c-v*s)[(uc)^2+uvcs+(vs)^2]=> u*c=v*stanQ=u/v => Q=atan(u/v) => cscQ=√(u^2+v^2)/v, secQ=√(u^2+v^2)/umin=y[atan(u/v)]=a*√(u^2+v^2)/v+b*√(u^2+v^2)/u=(u^3/v+v^3/u)√(u^2+v^2)=(u^4+v^4)√(u^2+v^2)/uv=[a^(4/3)+b^(4/3)]*[a^(1/3)+b^(1/3)]/(ab)^(1/3)max=y(π/2)=-a*0/1+b*1/0=∞