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I tried manipulating the terms but I couldn't get anywhere.

The only other thing I can observe is that the minimum must be greater than $1$ since all the terms are non-negative and the range of $\sec^4 x$ is $[1, \infty)$. Also it can't be $1$ since then $\sin^4 x + \cos^4$ must be $0$ which means $\sin x = \cos x = 0$ which is untrue for all $x$.

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Have you considered trying to take derivatives and seeing if there are any zeros? –  JB King Mar 22 '13 at 16:04
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3 Answers

up vote 6 down vote accepted

Hint: If you write $\sin^4 x$ as $(1-\cos^2 x)^2$ and $\sec x=\frac{1}{\cos x}$, then what you have is just a function of $\cos^2 x$, so you can set $u=\cos^2 x$ and look for a minimum of $(1-u)^2+u^2+u^{-2}$ under the condition that $u\in[0,1]$.

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For all non-negative $a$, we have the inequality $$\frac{1}{a}+a\geq 2.$$ We can deduce this from the fact that $(a-1)^2\geq 0\Rightarrow a^2+1\geq 2a$, and then dividing both sides by $a$. This inequality implies that $$\sin^4(x)+\cos^4(x)+\sec^4(x)\geq \cos^4(x)+\sec^4(x)\geq 2.$$

By considering $x=0$, we see that this lower bound is in fact optimal.

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$\sin^{4}{x}+\cos^{4}{x}+\sec^{4}{x}=\sin^{4}{x}+\left(\cos^{2}{x}-\sec^{2}{x}\right)^2+2$

Minimum value of $\sin^4{x}$ is zero and is attained at $x=n\pi$

The minimum value of $\cos^{2}{x}-\sec^{2}{x}$ is zero and attained when $\cos^{2}{x}=\sec^{2}{x}$ or $x=n\pi$. Both minimum's coincide at $x=n\pi$. And the minimum of the expression is $2$

Note: If the region where minimum is attained had not coincided , the calculus methods would be the best try.

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