Show that for all $x\in ]0,1[\;,\;2^{1-n} \leq x^n + (1-x)^n.$ Problem:
Show that for all $x\in ]0,1[$ and $n\in \mathbb{N}^*$ $$2^{1-n} \leq x^n + (1-x)^n$$
I've tried AM-GM inequality and induction, but both lead nowhere.
I have to show it without study of variations.
 A: Using induction, let $x,y \in (0,1)$ such that $x+y=1$ we want to show that for $n\in \mathbb{N}^*$
$$P_n:\frac{2}{2^n} \leq x^n + y^n$$
Note that:
$$x^n+y^n = (x^n+y^n)(\underbrace{x+y}_{=1})=x^{n+1}+y^{n+1}+yx^n+xy^n \;\;\;\;(*)$$
Assume without loss of generality that $y\geq x$ then
$$x^{n+1}+y^{n+1}-xy^n-yx^n=(y^n-x^n)(y-x)\geq 0 $$
Thus
$$x^{n+1}+y^{n+1} \geq yx^n+xy^n$$
Return to $(*)$
$$x^n+y^n = x^{n+1}+y^{n+1}+yx^n+xy^n\leq 2(x^{n+1}+y^{n+1})$$
Then
$$P_{n+1}: \frac{2}{2^{n+1}} = \frac{1}{2} \frac{2}{2^n} \leq \frac{x^n+y^n}{2} \leq x^{n+1}+y^{n+1}$$
A: It is equivalent to showing that for any $a \in [0,.5)$
$\;\;2^{1-n} \leq (.5-a)^n + (.5+a)^n$
But, applying the binomial theorem on $(.5-a)^n$ and $(.5+a)^n$ we see that
$\;\; (.5-a)^n + (.5+a)^n \ge (.5)^n +(.5)^n = 2 \cdot (2)^{-n} = 2^{1-n}$
A: You could use AM-RM inequality :
$\dfrac{a+b}2\leqslant\sqrt[n]{\dfrac{a^n+b^n}2}\qquad\forall\,a,b\in\mathbb R_0^+\;,\;\forall\,n\in\mathbb N\;.$
If $\;a=x\;$ and $\;b=1-x\;,\;$ we get that
$\dfrac12\leqslant\sqrt[n]{\dfrac{x^n+(1-x)^n}2}\qquad\forall\,x\in\big]0,1\big[\;,\;\forall\,n\in\mathbb N\;\;,$
$\dfrac1{2^n}\leqslant\dfrac{x^n+(1-x)^n}2\qquad\forall\,x\in\big]0,1\big[\;,\;\forall\,n\in\mathbb N\;\;,$
$2^{1-n}\leqslant x^n+(1-x)^n\qquad\forall\,x\in\big]0,1\big[\;,\;\forall\,n\in\mathbb N\;\;.$

Now, we will prove AM-RM inequality.
For any $\,a,b\in\mathbb R_0^+\;$ and for any $\,n\in\mathbb N\;,\;$ we get that
$\begin{align}
\dfrac{a+b}2&=\sqrt[n]{\left(\dfrac{a+b}2\right)^{\!n}}\leqslant\\
&\leqslant\sqrt[n]{\dfrac12\left[\left(\dfrac{a\!+\!b}2\!+\!\dfrac{a\!-\!b}2\right)^{\!n}\!+\!\left(\dfrac{a\!+\!b}2\!-\!\dfrac{a\!-\!b}2\right)^{\!n}\right]}=\\
&=\sqrt[n]{\dfrac{a^n+b^n}2}
\end{align}$
indeed, by applying the binomial theorem, it results that
$\left(\dfrac{a\!+\!b}2\!+\!\dfrac{a\!-\!b}2\right)^{\!n}\!+\!\left(\dfrac{a\!+\!b}2\!-\!\dfrac{a\!-\!b}2\right)^{\!n}\geqslant2\left(\dfrac{a+b}2\right)^{\!n}\,.$
A: $f_n(x) = x^n + (1-x)^n$ is convex on $[0,1]$ and symmetric about $x=1/2$, so its minimum is at $x=1/2$.  What is its value there?
