If a and b are positive, use mathematical induction to prove that `((a+b)/(2))^(n) le (a^(n)+b^(n))/(2) AA n in N`

Explanation

`P(n): ((a+b)/(2))^(n) le (a^(n)+b^(n))/(2)`
STEP-1: For n=1
`P(i): (a+b)/(2) le (a+b)/(2)`
Which is true for n=1
STEP-II: Assume it is true for n=k, then
`P(k): ((a+b)/(2))^(k)le (a^(k)+b^(k))/(2)`
STEP III: For n=k+1
`((a+b)/(2))^(k+1)=((a+b)/(2))((a+b)/(2))^(k)le ((a+b)/(2))((a^(k)+b^(k))/(2)) ("By assumption step")`
`Rightarrow ((a+b)/(2))^(k+1) le (1)/(4)[(a^(k+1)+b^(k+1))+a^(k)b+ab^(k)]`
`Rightarrow ((a+b)^(2)/2)^(k+1) le (1)/(4)[2(a^(k+1)+b^(k+1))-a^(k+1)=b^(k+1)+a^(k)b+ab^(k)]`
`Rightarrow ((a+b)^(2)/2)^(k+1) le (1)/(4)[2(a^(k+1)+b^(k+1))-(a-b)(a^(k)-b^(k))]`
`le ((a^(k+1)+b^(k+1))/(2))-((a-b)(a^(k)-b^(k)))/(4)`
`therefore ((a+b)/(2))^(k+1) le (a^(k+1)+b^(k+1))/(2)`
Hence the statement is true for n=k+1 any by the principle of induction it is true for all `n in N`.