I want to find the minimum of $f(x)=\dfrac{2}{\sin(x)}+\dfrac{3}{\cos(x)}$ when $0<x<\pi/2$ using the C-S inequality.
By using derivatives it is easy to show that the minimum is
$$(2^{2/3}+3^{2/3})^{3/2}$$
but i have no idea how to prove this using C-S ineq,my attempts ends with an inequality of opposite direction.
Best Answer
Theorem Let $a,\,b>0$. For $x\in(0,\,\pi/2)$, $(a^3\csc x+b^3\sec x)^2\ge(a^2+b^2)^3$, with equality iff $\tan x=a/b$.
Comment The present problem is $a=2^{1/3},\,b=3^{1/3}$.
Proof by Cauchy-Schwarz Use $u^2\ge\frac{(u\cdot v)^2}{v^2}$ with$$u=\left(\begin{array}{c} a^{3/2}\csc^{1/2}x\\ b^{3/2}\sec^{1/2}x \end{array}\right),\,v=\left(\begin{array}{c} a^{1/2}\sin^{1/2}x\\ b^{1/2}\cos^{1/2}x \end{array}\right),$$which are parallel iff $\tan x=a/b$. Then$$v^2=a\sin x+b\cos x=\sqrt{a^2+b^2}\cos(x-\arctan(a/b))\le\sqrt{a^2+b^2},$$which again is saturated at $\tan x=a/b$. I'll leave you to finish the calculation counting powers of $a^2+b^2$.