Find minimum of $\sin^4 x + \cos^4 x + \sec^4 x$ 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 $x$.

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