Trigonometry Minimum value using A.M. and G.M.

Question

If $f(x)=(\sin x + \csc x)^2+(\cos x + \sec x)^2$, then what is the minimum value of $f(x)$?

My Try : Now upon using Arithmetic Mean is greater than or equal to Geometric Mean I got minimum values for the following expressions as :-

$\sin^2x+\csc^2x=2$ and also, $\cos^2x+\sec^2x=2$

But surprisingly, the answer I get after opening the square and using this is $8$ while my book shows its $9$. Someone please tell me what I'm doing wrong.

Answer 1

HInt :$$f(x)=(\sin x + \csc x)^2+(\cos x + \sec x)^2\\=(\sin x +\frac{1}{\sin x})^2+(\cos x +\frac{1}{\cos x})^2=$$ $$\sin^2x+\frac{1}{\sin^2 x}+2+\cos ^2 x+\frac{1}{\cos^2 x}+2=\\ 5+\frac{1}{\sin^2 x}+\frac{1}{\cos^2 x}=\\ 5+\frac{{\sin^2 x+\cos^2 x}}{\sin^2 x\cos^2 x}=\\ 5+\frac{1}{\sin^2 x\cos^2 x}$$now take $$a=\sin^2x, b=\cos^2x \\a+b=1 \\a+b \geq 2\sqrt{ab} \to \sqrt{ab}\leq \frac{a+b}{2}=\frac12 \to \max(ab)=\frac{1}{2^2}\\ \min\left(\frac{1}{\sin^2 x\cos^2 x}\right)=\min\left(\frac{1}{ab}\right)=\frac{1}{\max({ab})}=\frac{1}{\frac{1}{4}}=4$$ finally $$\min(f)=5+4$$

Answer 2

If you really want to use AM-GM inequality, Khosrotash's answer is enough. But...

Using Cauchy–Schwarz inequality:

$$(\sin x + \csc x)^2+(\cos x + \sec x)^2\\=[(\sin x + \csc x)^2+(\cos x + \sec x)^2][\sin^2 x+ \cos^2 x]\\\geq \mid (\sin x + \csc x)\sin x+(\cos x + \sec x)\cos x \mid^2\\=\mid (\sin^2 x +1)+(\cos^2 x +1) \mid^2\\=\mid 1+1+1 \mid^2=9\space\space\space\space\blacksquare$$

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To prove $(a^2+b^2)(c^2+d^2)\geq |ac+bd|^2$, you need to do this: $$(a^2+b^2)(c^2+d^2)\\=|ac|^2+|bd|^2+(|ad|^2+|bc|^2)$$

$\dfrac{|ad|^2+|bc|^2}{2}\geq |ad|\cdot|bc|=|ac|\cdot|bd|$( by AM-GM inequality).

Hence $$(a^2+b^2)(c^2+d^2)\\\geq |ac|^2+|bd|^2+2|ac|\cdot|bd|\\=(|ac|+|bd|)^2\\\geq |ac+bd|^2\space\text{(by Triangle Inequality).}$$

Answer 3

How about this:

$f(x) = (\sin x + \csc x)^2 + (\cos x + \sec x)^2$.

Simplifying this yields:

$f(x) = 5 + \frac{1}{\sin^2(x)\cos^2(x)}$.

Note : $ \sin^2(x) \cos^2(x) =$

$(1/4) [4\sin^2(x) \cos^2(x)] =$

$(1/4)[\sin(2x)]^2$.

$f(x) = 5 + \frac{4}{[\sin(2x)]^2}$.

For $ \sin(2x) = 1$ we find

the minimum value:

$\min (f(x)) = 5 + 4 = 9$.