A proof of fraction inequality by induction Prove that $$\left(\frac{a+b}{2}\right)^n\le\frac{a^{n}+b^{n}}{2}$$ I have checked the base case and assumed the inequality is valid and continued like this $$\left(\frac{a+b}{2}\right)^{n+1}\le\frac{a^{n+1}+b^{n+1}}{2}\Rightarrow \left(\frac{a+b}{2}\right)^n\left(\frac{a+b}{2}\right)\le\frac{a^{n+1}+b^{n+1}}{2}$$ Multiplying by $2$ I got $$\frac{(a+b)^{n+1}}{2^n}\le a^{n+1}+b^{n+1}$$ and I don't know how to continue from here. I've also tried using the binomial theorem to get something different but couldn't get anything useful.
 A: Work:
$\left( \dfrac{a+b}{2}\right)^{n+1}=\left(\dfrac{a+b}{2}\right)^n\left(\dfrac{a+b}{2}\right)$
$ \hspace{2.6  cm}\leq \dfrac{a^n+b^n}{2}\cdot \left(\dfrac{a+b}{2}\right)$
$\hspace{2.6 cm} = \dfrac{a^{n+1}+a^nb+ab^n+b^{n+1}}{4}$
Now write what you want to obtain and subtract extras:
$\hspace{2.6 cm} = \dfrac{a^{n+1}+b^{n+1}}{2}-\dfrac{a^{n+1}+b^{n+1}}{4}+\dfrac{a^nb+ab^n}{4}$
$\hspace{2.6cm} = \dfrac{a^{n+1}+b^{n+1}}{2}-\left(\dfrac{a^{n+1}-a^nb-ab^n-b^{n+1}}{4}\right)$
$\hspace{2.6cm} = \dfrac{a^{n+1}+b^{n+1}}{2}-\left(\dfrac{a^{n}(a-b)-b^n(a-b)}{4}\right)$
$\hspace{2.6cm} = \dfrac{a^{n+1}+b^{n+1}}{2}-\dfrac{(a-b)(a^n-b^n)}{4}=*$
No matter if $a\geq b$ or $b> a$ expression $ \dfrac{(a-b)(a^n-b^n)}{4}$ is positive. So, in conclusion:
$\hspace{2.6cm}*\leq \dfrac{a^{n+1}+b^{n+1}}{2}$
A: It's just $$\frac{a^{n+1}+b^{n+1}}{2}\geq\frac{a+b}{2}\cdot\frac{a^n+b^n}{2}\geq\left(\frac{a+b}{2}\right)^{n+1},$$
here the left inequality it's $(a^n-b^n)(a-b)\geq0$, which is obvious and
the right inequality it's a hypothesis of the induction. 
