Mathematical Induction: Prove [(a+b)/2]^n < (a^n+b^n)/2

[bbq999]

Question: Prove, by induction, that:

[(a+b)/2]^n < (a^n+b^n)/2,

where a>b>0 and n = 2, 3, 4, ...

Can anybody help me? Thanks in advance.

 

[pka]

Have you shown it is true for n=2?
If please post your effort.
If not, tell us what trouble you are having.

 

[soroban]

Re: Mathematical Induction: Prove [(a+b)/2]^n < (a^n+b^n)

Hello, bbq999!

pka's concern is quite justified.

Are you familiar with the Induction procedure?
If so, which stage is giving you trouble?

Prove, by induction, that: $\displaystyle \:\left(\frac{a+b}{2}\right)^n\:<\:\frac{a^n+b^n}{2}\:$ where $\displaystyle a\,>\,b\,>\,0$ and $\displaystyle n \,= \,2,\,3,\,4,\,...$

Even the initial case is tricky to verify . . .

For $\displaystyle n\,=\,2:$
. . The left side is: $\displaystyle \:\left(\frac{a+b}{2}\right)^2 \:=\:\frac{1}{4}\left(a^2\,+\,2ab\,+\,b^2\right)$
. . The right side is: $\displaystyle \:\frac{a^2\,+\,b^2}{2}$
Is the left side smaller?


We have: \(\displaystyle \:\frac{a^2\,+\,2ab\,+\,b^2}{4}\;\;{\begin{array}{c}<\\>\end{array} \;\;\frac{a^2\,+\,b^2}{2}\)

Multiply by 4: $\displaystyle \:a^2\,+\,2ab\,+\,b^2\;\;\begin{array}{c}<\\>\end{array}\;\;2a^2\,+\,2b^2$

Then we have: $\displaystyle \:0 \;\;\begin{array}{c}<\\>\end{array}\;\;a^2\,-\,2ab\,+\,b^2$

. . Hence: \(\displaystyle \:0\;\;\begin{array}<\\>\end{array}\;\;(a\,-\,b)^2\)


For $\displaystyle a\,\neq\,b$, the right side is positive.

. . Therefore, the inequality is less than: \(\displaystyle \;\L<\)

We have verified the initial case . . . whew!


Can you finish the proof now?