Pure Maths 二項式

Consider a>0 ,

(a-1)(a^k-1)>0 -> a+a^k<=1+a^(k+1)

let P(n) is "(1+a)^n<=2^(n-1) (1+a^n)"

when n=1 , (1+a)^1 = 2^(1-1) (1+a^1)

it is true for n=1 ,

Assume it is true for n=k , where k is a positive integer ,

i.e. (1+a)^k<=2^(k-1)(1+a^k)

Consider n=k+1 ,

(1+a)^k+1

=(1+a)^k (1+a)

<=2^(k-1)(1+a^k) (1+a)

<=2^(k-1)(1+a+a^k+a^(k+1))

=2^(k)(1+a^k+1)//

it is true for n=k+1 ,

By MI, it is true for all positive integer n.

Put a=x/y ,

(1+x/y)^n<=2^(n-1)(1+(x/y)^n) -> [(x+y)/2]^n<=(x^n+y^n)/2

前面既做法只係「induction on n」，冇假設a, b係整數。

除左MI，如果你有學Convex Function既話，
就可以話因為 f(x) = x^m 係convex function，
所以 f( λa + (1-λ)b ) <= λf(a) + (1-λ) f(b)，
取 λ=1/2，咁題目既不等式就搞掂。