已知函数f(x)=e的X次方减kx,x属于R

Proof. First of all, F(x)=e^x+e^{-x}. We shall handle it according to the parity of n.

If n is even, then F(1)F(2)...F(n)=[F(1)F(n)][F(2)F(n-1)]...[F(n/2)F(n/2+1)]. (1)
For any 1<=j<=n/2, we have
F(j)F(n+1-j)
= (e^j+e^{-j}) (e^{n+1-j}+e^{-n-1+j})
= e^{n+1} + (e^{n+1-2j}+e^{2j-n-1}) + e^{-n-1}. (2)
Since the second summand in the above expression (2) is a sum of the form t+t^{-1}, having minimum 2, while the third summand e^{-n-1} is positive, we deduce F(j)F(n+1-j)>e^{n+1}+2. It follows from (1) that
F(1)F(2)...F(n)>(e^{n+1}+2)^{n/2}.

If n is odd, then F(1)F(2)...F(n)=[F(1)F(n)][F(2)F(n-1)]...[F((n-1)/2)F((n+1)/2+1)]F((n+1)/2). (3)
Similar to the case n is even, the formula (2) holds true for any 1<=j<=(n+1)/2, which is larger than e^{n+1}+2. It follows from (3) that
F(1)F(2)...F(n)>(e^{n+1}+2)^{(n-1)/2}(e^{n+1}+2)^{1/2}=(e^{n+1}+2)^{n/2}.

This completes the proof.