Prove that $\sin^n (2x) + (\sin^n x - \cos^n x)^2 \leq 1$ 
Prove that $\sin^n (2x) + (\sin^n x - \cos^n x)^2 \leq 1$.

Let $a = \sin x, b = \cos x.$ Then we want to show $2^n a^n b^n +a^{2n}-2a^n b^n + b^{2n}\leq 1,$ which follows once we show that $(2^n-2)a^n b^n \leq \sum_{j=1}^{n-1}{n\choose j} a^{2(n-j)}b^{2j}$. I think the AM-GM inequality is useful, but I'm not sure how to apply it in this case. I don't think that for all j we have $a^{2(n-j)} b^{2j}\ge a^n b^n$; for instance if $a<b,$ then $a^{2(n-j)}b^{2j} < a^nb^n$.
 A: Assume that $n$ is a positive integer.
Let $a = \sin x$ and $b = \cos x$.
Then $a^2 + b^2 = 1$ and $|2ab| \le a^2 + b^2 = 1$.
Let $$f(n) := (2^n-2) a^n b^n + a^{2n} + b^{2n}.$$
We have $f(1) = a^2 + b^2 = 1$ and $f(2) = 2a^2b^2 + a^4 + b^4 = (a^2 + b^2)^2 = 1$.
First, we have
\begin{align*}
 &f(2n) - f(2n+1)\\
 =\,& (2^{2n}-2) a^{2n} b^{2n} + a^{4n} + b^{4n}
 - [(2^{2n+1}-2) a^{2n+1} b^{2n+1} + a^{4n+2} + b^{4n+2}]\\
 =\,& (2^{2n}-2)(1 - 2ab)a^{2n}b^{2n} - 2a^{2n+1}b^{2n+1} + a^{4n}(1 - a^2) + b^{4n}(1 - b^2)\\
 \ge\,& - 2a^{2n+1}b^{2n+1} + a^{4n}b^2 + b^{4n}a^2\\
 \ge\,& - 2a^{2n+1}b^{2n+1} + 2\sqrt{a^{4n}b^2 \cdot b^{4n}a^2}\\
 \ge\,& 0.
\end{align*}
Second, we have
\begin{align*}
 &f(2n) - f(2n+2)\\
 =\,& (2^{2n}-2) a^{2n} b^{2n} + a^{4n} + b^{4n} - [(2^{2n+2}-2) a^{2n+2} b^{2n+2} + a^{4n+4} + b^{4n+4}]\\
 =\,& (2^{2n}-2)(1-4a^2b^2)a^{2n}b^{2n} - 6a^{2n+2}b^{2n+2} + a^{4n}(1-a^4) + b^{4n}(1-b^4)\\
 \ge\,& - 6a^{2n+2}b^{2n+2} + a^{4n}b^2(1+a^2) + b^{4n}a^2(1+b^2)\\
 =\,& a^2b^2(a^{4n} + b^{4n} - 2a^{2n}b^{2n}) + a^{4n}b^2 + b^{4n}a^2 - 4a^{2n+2}b^{2n+2}\\
 \ge\,& 2\sqrt{a^{4n} b^2 b^{4n} a^2} - 2ab \cdot 2a^{2n+1}b^{2n+1}\\
 \ge\,& 0.
\end{align*}
We are done.
A: For $\cos x=0$ the inequality holds and for $\cos x\neq0$ by $t=\tan x$ we have
$$\sin^n (2x) + (\sin^n x - \cos^n x)^2 \leq 1 \iff 2^n t^n+(t^n-1)^2\le(1+t^2)^n$$
$$\begin{aligned}
&\iff 2^n t^n+t^{2n}-2t^n+1\le\sum_{k=0}^{n}\binom n k t^{2k}=t^{2n}+1+\sum_{k=1}^{n-1}\binom n k t^{2k}\\
&\iff (2^n-2) t^n\le\sum_{k=1}^{n-1}\binom n k t^{2k}\\
&\iff \frac{\sum_{k=1}^{n-1}\binom n k t^{2k}}{2^n-2}\ge t^n
\end{aligned}$$
and assuming WLOG $\;t>0$, the latter is satisfied by weighted AM-GM inequality, indeed by

*

*$\sum_{k=1}^{n-1}\binom n k =2^n-2$

*$\sum_{k=1}^{n-1}k\binom n k =\frac12 n(2^n-2)$
we obtain
$$\frac{\sum_{k=1}^{n-1}\binom n k t^{2k}}{\sum_{k=1}^{n-1}\binom n k} \ge t^{\left(\frac{\sum_{k=1}^{n-1}2k\binom n k}{\sum_{k=1}^{n-1}\binom n k}\right)}=t^n$$
