Show $\forall x \in \mathbb{R}^* \quad (1+x)^n \leq 2^{n-1}(1+x^n)$ Let $n\in N$ such that $n \geq 2$
$\forall x \in \mathbb{R}^*  \quad (1+x)^n \leq 2^{n-1}(1+x^n)$ 
I tried induction but with no luck. On the other hand I said that $f(x)=x^n$ is convex then by Jenson inquality for $f$ with $ f((a+b)/2) \leq (f(a)+f(b))/2 $.
 A: By induction for $x \in [0,+\infty)$: start with $n=2$. You know that in general $(x-1)^2 \geq 0$. This means $x^2-2x+1 \geq 0$, so $2x \leq 1+x^2$, and adding $1+x^2$ to both sides you get
$$(1+x)^2 = 1+2x+x^2 \leq 2(1+x^2)$$
which is your base case. Now assume the formula works for $n\geq2$. Then 
$$(1+x)^{n+1} = (1+x)(1+x)^n \leq 2^{n-1}(1+x)(1+x^n).$$
Now you want to prove that $2^{n-1}(1+x)(1+x^n) \leq 2^n(1+x^{n+1})$, which is equivalent to $1+x+x^n+x^{n+1} \leq 2+2x^{n+1}$, namely $x^{n+1} -x^n-x+1 \geq 0$. This is true since $x \geq 0$ and
$$x^{n+1} -x^n-x+1 = (1-x)-x^n(1-x) = (1-x^n)(1-x) = (1-x)^2(1+x+\dots+x^{n-1}) \geq 0.$$
A: If one knows that $f\colon \, [0,\infty)\longrightarrow \mathbb{R}$ is a convex function then one has
$$
f\left(t x_1+(1-t)x_2\right)\le tf\left(x_1\right)+(1-t)f\left(x_2\right),\qquad t \in [0,1],
$$ giving, with $f=x^n,x_1=1,x_2=x,t=\frac12$, $n\ge2$
$$
\left(\frac{1+x}{2}\right)^n\le \frac{1}{2}+\frac{x^n}{2},\qquad x \in [0,\infty),
$$ which is equivalent to the expected inequality.
