pure (applications of diff.)

f'(x)=[(s-x)/k]^k - x[(s-x)/k]^(k-1)
Set f'(x)=0,x=s/(k+1) and f(x)=[s/(k+1)]^(k+1)
(b) f(x)=x[(s-x)/k]^k
when n=1, the statement is obviously true.
Assume that when n=k, the statement is true. i.e.
(a1+a2+...+an)/n>=(a1a2...an)^(1/n)
when n=k+1
Sub. s=a1+a2+...+an+1,x=an+1,k=n
f(x)=x[(s-x)/k]^k=(an+1)[(a1+a2+...+an)/n]^n
Since f(x) attains its max. at s/(k+1).
So when an+1=(a1+a2+...+an)/n, f(an+1) attains its maximum
[(a1+a2+...+an+an+1)/(n+1)]^(n+1)
Since f(an+1)=(an+1)[(a1+a2+...+an)/n]^n
This implies that
[(a1+a2+...+an+an+1)/(n+1)]^(n+1) >=(an+1)[(a1+a2+...+an)/n]^n
>=(an+1)(a1a2...an)
Equality holds when an+1=(a1+a2+...+an)/n. However, since
a1=a2=...=an, so an+1=a1=a2=...=an
(c) First, consider how to obtains n+1
We know that a1+a2+...+an=1
So (1+a1)+(1+a2)+...+(1+an)=n+1 but the question is talking about 1-ai. So, how to change nimus to plus ? The key point is : (1-x)^(-1)=1+x+x^2+...+x^n+...
So (1-ai)^(-1)=1+ai+(ai)^2+...> 1+ai
Thus Σ 1/(1-ai) > Σ (1+ai) = n+1