Proving a trigonometric inequality I've been having difficulty with the following,
Prove that,
$[\sin^{n+1}(x)]^{2}+[\cos^{n+1}(x)]^{2} \geq (\frac12)^{n}$ where $x$ is real and $n$ is a non-negative integer.
I've tried an inductive approach but have struggled with the inductive step.
 A: use Holder inequality
$$[(\cos^2{x})^{n+1}+(\sin^2{x})^{n+1}][1+1]^{n} \geq (\cos^2{x}+\sin^2{x})^{n+1}=1$$
or
Use AM-GM inequality we have
$$(\sin^2{x})^{n+1}+\dfrac{1}{2^{n+1}}+\dfrac{1}{2^{n+1}}+\cdots+\dfrac{1}{2^{n+1}}\ge (n+1)\cdot\sqrt[n+1]{\dfrac{1}{2^{n(n+1)}}(\sin^2{x})^{n+1}}=\dfrac{n+1}{2^{n}}\sin^2{x}$$
so
$$(\sin^2{x})^{n+1}+\dfrac{n}{2^{n+1}}\ge\dfrac{n+1}{2^{n+1}}\sin^2{x}\tag{1}$$
the same as
$$(\cos^2{x})^{n+1}+\dfrac{n}{2^{n+1}}\ge\dfrac{n+1}{2^{n+1}}\cos^2{x}\tag{2}$$
$(1)+(2)$
$$\sin^{2n+2}{x}+\cos^{2n+2}{x}\ge\dfrac{1}{2^n}$$
A: We have $\displaystyle |\sin(x)|\geq \frac{1}{\sqrt {2 }}$ or $\displaystyle |\cos(x)|\geq \frac{1}{\sqrt {2 }}$. Suppose wlog $\displaystyle |\sin(x)|\geq \frac{1}{\sqrt {2 }}$, and put $\displaystyle u=(\sin(x))^2\in [\frac{1}{2},1]=I$. We have then to show that $f(u)=u^{n+1}+(1-u)^{n+1}\geq \frac{1}{2^n}$. The computation of the derivative of $f$ show that on $I$, $f(u)\geq f(1/2)=1/2^n$ and we are done.  
A: As you can check by derivation, the expression
$$t^{n+1}+(1-t)^{n+1}$$ has a unique minimum at $t=\frac12$. 
The rest easily follows ($t=\sin^2(x)$).
A: You may also use power means inequality to generalise the statement to all positive real $n$.  Take $x_1=\sin^2x,x_2=\cos^2x$.  Observe that $x_1,x_2\ge0$, and
\begin{align}
\left(\frac{x_1^{n+1}+x_2^{n+1}}{2}\right)^\frac1{n+1} &\ge \frac{x_1+x_2}{2} \\
\left(\frac{\sin^{2n+2}x+\cos^{2n+2}x}{2}\right)^\frac1{n+1} &\ge \frac{\sin^2x+\cos^2x}{2} = \frac12 \\
\sin^{2n+2}{x}+\cos^{2n+2}{x}&\ge\dfrac{1}{2^n}.
\end{align}
