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While solving a problem I came across this task, minimizing \begin{align} \left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2. \end{align}

One can easily do it with calculus to show that the minimum value is $12.5$. I tried to do it using trigonometric identities and fundamental inequalities (like AM-GM, Cauchy-Schwarz, etc.) but failed. Can someone help me to do it using trig identities and inequalities?

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Knowing the answer it's not that difficult to get it another way. Let $y = \cos 2x$. We have $\sin^2 x = \frac{1-y}{2}$, $\cos^2 x =\frac{1+y}{2}$. Then \begin{align} \left(\sin^2 x + \frac{1}{\sin^2 x}\right)^2 + \left(\cos^2 x + \frac{1}{\cos^2 x}\right)^2 &= \left(\frac{1-y}{2} + \frac{2}{1-y}\right)^2 + \left(\frac{1+y}{2} + \frac{2}{1+y}\right)^2 = \\ &= \frac{y^6+7y^4-y^2+25}{2(1-y^2)^2} = \\ &= \frac{25}{2} + \frac{y^2(y^4-18y^2+49)}{2(1-y^2)^2} = \\ &= \frac{25}{2} + y^2\Big(\frac{1}{2} + \frac{8}{1-y^2} + \frac{16}{(1-y^2)^2}\Big) \end{align} Since $y^2 \le 1$, the expression in the brackets is strictly positive.

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The following is not completely rigourous, but can be made so with little effort.

An obvious trigonometric identity is $\sin^2 x+\cos^2 x=1$. Using this, put $y=\sin^2 x$ to obtain a function of the form

$$f(y)=(y+\frac{1}{y})^2+(1-y+\frac{1}{1-y})^2$$ defined on the open interval $(0,1)$. This function is symmetric around $y=\frac{1}{2}$, because $f(y)=f(1-y)$. Also, $\lim_{y\to 0^+}f(y)=\lim_{y\to 1^-}f(y)=+\infty$. You can use a bit of calculus to convince yourself that $f$ decreases in $(0,1/2)$ and increases in $(1/2,1)$ to conclude that there is a minimum (unique, actually), when $y=1/2$.

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Here is a solution using AM-GM and the double-angle formulae

  • $(\star):\sin^2 x = \frac{1-\cos 2x}{2}$, $\cos^2 x = \frac{1+\cos 2x}{2}$

\begin{eqnarray*}\left ( \sin^2 x + \frac{1}{\sin^2 x} \right )^2 + \left ( \cos^2x + \frac{1}{\cos^2x} \right )^2 & \stackrel{AM-GM}{\geq} & 2\left ( \sin^2x + \frac{1}{\sin^2x} \right ) \left ( \cos^2x+ \frac{1}{\cos^2x} \right ) \\ & = & 2\frac{(\sin^4 x + 1)(\cos^4 x + 1)}{\sin^2 x \cdot \cos^2 x}\\ & \stackrel{AM-GM}{\geq} & 8\frac{(\sin^4 x + 1)(\cos^4 x + 1)}{(\sin^2 x + \cos^2 x)^2}\\ & \stackrel{\star}{=} & 8\left(\frac{(1-\cos 2x)^2}{4}+1 \right)\left(\frac{(1+\cos 2x)^2}{4}+1 \right)\\ & = & \frac{1}{2}(5+\cos^2 2x - 2\cos 2x)(5+\cos^2 2x + 2\cos 2x)\\ & = & \frac{1}{2}(25+\cos^4 2x +6 \cos^2 2x)\\ & \geq & \frac{25}{2} \end{eqnarray*} Note that equality is attained for $\cos 2x = 0 \Leftrightarrow \cos^2 x = \sin ^2 x$, where the last condition is required to produce equality for AM-GM.

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Another method: $$\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2=\\ \sin ^4x+\cos ^4x+4+\frac1{\sin^4 x}+\frac1{\cos^4 x}=\\ (\sin ^2x+\cos ^2 x)^2-2\sin^2x\cos^2x+4+\frac{(\sin ^2x+\cos ^2 x)^2-2\sin^2x\cos^2x}{\sin^4x\cos^4x}=\\ 1-\frac12\sin^2 2x+4+\frac{1-\frac12\sin 2x}{\frac1{16}\sin^4 2x}=\\ \left(1-\frac12\sin^2 2x\right)\left(1+\frac{16}{\sin^4 2x}\right)+4\ge \\ \left(1-\frac12\cdot 1\right)\left(1+\frac{16}{1^2}\right)+4=12.5. $$ Note: $1-\frac12\sin ^2 2x>0$, so $\sin^2 2x$ should be maximized to $1$.

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AM-GM says $a^2+b^2\geq 2ab$ with equality when $a=b.$ Thus we cannot obtain a lower value than which gives GM.
With a bit of algebra we get that the equality really occurs, and this is iff $|\sin x|=|\cos x|,$ from where the minimum $$2.5^2+2.5^2=12.5$$

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