$\sin^{n}x+\cos^{n}x$

It is clear that for $n \leq 2$ the inequality is trivial so suppose that $n \geq 2$ . Then, by the power mean inequality:

$\sqrt[n]{\frac{\sin^n x + \cos^n x}{2}} \geq \sqrt{ \frac{\sin^2 x + \cos^2 x}{2}} = \sqrt{\frac{1}{2}}$ Re-arranging the important parts of this we get: $\sin^n x + \cos^n x \geq 2 \left(\frac{1}{2}\right)^{\frac{n}{2}}$

So if $n$ is an integer such that the inequality in the problem does not hold, it must be that $2 \left(\frac{1}{2}\right)^{\frac{n}{2}} < \frac{1}{n}$ . This inequality holds for $n \geq 9$ (check proof below). However, this condition is necessary but not sufficient so let's confirm that $n \geq 9$ does not work: Consider $x = \frac{\pi}{4}$ which implies $\sin x = \cos x = \frac{\sqrt{2}}{2}$ . Then $\sin ^n x + \cos ^n x = 2\left( \frac{\sqrt{2}}{2} \right)^n = 2 \left(\frac{1}{2}\right)^{\frac{n}{2}} < \frac{1}{n}$ .

Thus, the given inequality holds up to $n=8$ .

$2 \left(\frac{1}{2}\right)^{\frac{n}{2}} < \frac{1}{n} \iff 0.5^{\frac{n}{2}} < \frac{0.5}{n}$ . Let $u = \frac{n}{2}$ , then $0.5^{\frac{n}{2}} < \frac{0.5}{n} \iff \left( \frac{1}{2} \right)^u < \frac{1}{4u} \iff -u \log 2 < - \log 4u \iff \frac{\log 4u}{u \log 2} < 1 \iff \frac{2}{u} + \frac{\log_2 u}{u} < 1$ . $\frac{2}{u}$ is a decreasing function for $u > 0$ and it is a trivial calculus exercise to prove that $\frac{\log_2 u}{u}$ is decreasing for $u > e$ . Thus overall the expression is decreasing and checking for $u = 4.5$ (which corresponds to $n=9$ ) we see that indeed $\frac{2}{u} + \frac{\log_2 u}{u} < 1$ and as it is decreasing, it must hold for all $u > 4.5$ too.

Note that $n$ can't be odd, otherwise at $x = \pi$ we would have $\sin^n x + \cos^n x = 0^n + (-1)^n = -1 < \frac1n$ . Hence we might assume that $n$ is even.

Let $f_n(x) = \sin^n x + \cos^n x.$ Our goal is to find the minimum of $f_n$ . Computing the derivative yields $f_n(x)' = n\sin^{n-1} x \cos x + n\cos^{n-1} x (-\sin x) = n(\sin^{n-1} x \cos x - \cos^{n-1} x \sin x )$ . Clearly the function can only have minimum value where its derivative equals zero, therefore we get the equation

$n(\sin^{n-1} x \cos x - \cos^{n-1} x \sin x) = 0$

$\sin^{n-1} x \cos x - \cos^{n-1} x \sin x = 0$

$\sin x \cos x (\sin^{n-2} x - \cos^{n-2} x) = 0$

This can only be satisfied when either $\sin x = 0$ or $\cos x = 0$ , which implies $x = k\frac{\pi}{2}$ for $k \in \mathbb{Z}$ , or alternatively

$\sin^{n-2} x = \cos^{n-2} x$

$\tan^{n-2} x = 1,$

where we assumed that $\cos x \neq 0.$ This equation has solutions $\tan x = \pm 1$ or $x = \frac{\pi}{4} + k\frac{\pi}{2}$ , since $n$ is even.

Note that in the former case of $x = k\frac{\pi}{2}$ one of $\sin x$ and $\cos x$ is zero, and the other has absolute value of $1$ . Therefore $\sin^n x + \cos^n x = 0^n + (\pm 1)^n = 1 \geq \frac1n$ . However, in the latter case of $x = \frac{\pi}{4} + k\frac{\pi}{2}$ the absolute value of both $\sin x$ and $\cos x$ equals to $\frac{1}{\sqrt{2}}$ , hence $\sin^n x + \cos^n x = (\pm \frac{1}{\sqrt{2}})^n + (\pm \frac{1}{\sqrt{2}})^n = 2 \cdot 2^{-\frac{n}{2}} = 2^{1-\frac{n}{2}}.$ It follows that the minimum value of $f_n$ is either $1$ or $2^{1-\frac{n}{2}}$ .

In the first case the condition is clearly satisfied. However, in the second case $\sin^n x + \cos^n x \geq \frac1n$ only holds when $2^{1-\frac{n}{2}} \geq \frac1n$ or equivalently $2n \geq 2^{\frac{n}{2}}.$ Note that this is true whenever $n \leq 8$ . To show it doesn't hold for $n > 8$ we employ induction. The base case $n =9$ is clear. Assume that $2n < 2^{\frac{n}{2}}$ for some $n > 8$ . From the induction hyphothesis and the constraint $n > 8$ we get

$2^{\frac{n+1}{2}} = \sqrt{2} \cdot 2^{\frac{n}{2}} > \sqrt{2} \cdot 2n = 2n + 2(\sqrt{2}-1)n > 2n + 2(\sqrt{2}-1)8 > 2n+2 = 2(n+1),$

since $128 > 81 \Rightarrow 8\sqrt{2} > 9 \Rightarrow \sqrt{2}-1 > \frac18 \Rightarrow 16(\sqrt{2}-1) > 2.$ The induction is therefore complete.

We have obtained that the condition of the problem only holds for even $n \leq 8$ . The greatest such $n$ is $8$ , hence the solution.