Question

Cho \(n\in N\)và \(a+b>0\). Chứng minh rằng \(\left(\dfrac{a+b}{2}\right)^n\le\dfrac{a^n+b^n}{2}\)

Answer 1

Use Be-loli 's ineq:

\(\left(\dfrac{2a}{a+b}\right)^n=\left(1+\dfrac{a-b}{a+b}\right)^n\ge1+\dfrac{n\left(a-b\right)}{a+b}\)

\(\left(\dfrac{2b}{a+b}\right)^n=\left(1-\dfrac{a-b}{a+b}\right)^n\ge1-\dfrac{n\left(a-b\right)}{a+b}\)

Cộng theo vế 2 BĐT trên ta có:

\(\left(\dfrac{2a}{a+b}\right)^n+\left(\dfrac{2b}{a+b}\right)^n\ge2\Leftrightarrow\left(\dfrac{a+b}{2}\right)^n\le\dfrac{a^n+b^n}{2}\)