How to prove that $(1+x)^{2n}>1+2nx$? How to prove that $(1+x)^{2n}>1+2nx$ , $x\neq 0$  using induction?
Any hint would be appreciated.
 A: If $x>0$, by the induction hypothesis:
$$(1+x)^{2n}=(1+x)^2(1+x)^{2n-2}>(1+x)^2(1+2nx - 2x)$$
$$=1+2nx + ((2 n -2)x^3 + (4 n - 3)x^2)>1+2nx$$
Q.E.D.
A: Here is my way. Let $f(x)=(1+x)^{2n}-2nx$ on $[0,\infty)$.You can easily show that $f$ is increasing.(see what happens to $f'$??).Therefore for $t>0$ we obtain $1=f(0)<f(t)=(1+t)^{2n}-2nt$ or equivalently $(1+t)^{2n}>1+2nt$.
Similar way you can show that inequality holds for $x<0$
A: Base case:
$$(1+x)^{2\cdot1}=1+2x+x^2>1+2\cdot1\cdot x$$
Induction Hypothesis:
Assume that
$$(1+x)^{2k}>1+2kx$$
Induction step:
$$
\begin{align}
(1+x)^{2(k+1)}&=(1+x)^{2k}\cdot(1+x)^2 \\
&>(1+x)^{2k}\cdot(1+2x) \tag{By the Base Case} \\
&>(1+2kx)(1+2x) \tag{By the Induction Hypothesis} \\
&=1+2x+2kx+4kx^2 \\
&>1+2x+2kx \\
&>1+2x(k+1) \\
&\text{QED}
\end{align}
$$
A: The 2 years old answers are somewhat incomplete regarding the negative case so here we go: $f(x)=(1+x)^{2n}$ with $n\geq 1$ has as second derivative:
  $$ f''(x) = 2n (2n-1) (1+x)^{2(n-1)} $$
which is strictly positive for $x\neq 0$. $f$ is therefore strictly convex whence strictly above its tangents except at the touching point. And $f(0)+f'(0)x = 1 + 2n x$ is the tangent at $(0,1)$.
