Find the minimum value of $(\sin x+ \csc x)^2 + (\sec x + \cos x)^2$

Question

I made it till $5+\tan^2x + \cot^2x$, but I don’t know how to proceed further.

Answer 1

Now, by AM-GM $$5+\tan^2x+\cot^2x\geq5+2=7.$$ The equality occurs for $\tan^2=\cot^2x,$ which says that we got a minimal value.

But for your problem we can use C-S: $$\left(\sin{x}+\frac{1}{\sin{x}}\right)^2+\left(\cos{x}+\frac{1}{\cos{x}}\right)^2=5+\frac{1}{\sin^2x}+\frac{1}{\cos^2x}=$$ $$=5+(\sin^2x+\cos^2x)\left(\frac{1}{\sin^2x}+\frac{1}{\cos^2x}\right)\geq5+2^2=9.$$ The equality occurs for $(\sin{x},\cos{x})||\left(\frac{1}{\sin{x}},\frac{1}{\cos{x}}\right),$ which says that $9$ is a minimal value.

In both problems for equality occurring we can take $x=45^{\circ}.$

In the first solution I used $$a^2+b^2\geq2ab,$$ which is just $$(a-b)^2\geq0,$$ for $a=\tan{x}$ and $b=\cot{x}.$

In the second solution I used $$(a^2+b^2)\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\geq4,$$ which is $$2+\frac{a^2}{b^2}+\frac{b^2}{a^2}\geq4$$ or $$a^4+b^4-2a^2b^2\geq0$$ or $$(a^2-b^2)^2\geq0,$$ which is obvious.

Answer 2

$$\tan^2x+\cot^2x=(\cot x-\tan x)^2+2\cot x\tan x\ge2$$ for real $\cot x-\tan x$

Answer 3

I don't think your function is $5+\tan^2x+\cot^2x$, actually. With $c:=\cos x,\,s:=\sin x$ your sum is $$\left(s+\frac{1}{s}\right)^2+\left(c+\frac{1}{c}\right)^2=5+\frac{1}{s^2}+\frac{1}{c^2}=5+\frac{1}{c^2s^2}=5+4\csc^22x\ge 9,$$with equality iff $\sin2x=\pm 1$, as occurs e.g. if $x=\frac{\pi}{4}$ for which we get$$\left(s+\frac{1}{s}\right)^2+\left(c+\frac{1}{c}\right)^2=2\left(\frac{1}{\sqrt{2}}+\sqrt{2}\right)^2=2\left(\frac{3}{\sqrt{2}}\right)^2=9.$$