Verified answer

Suppose x > -1 and let P[n] stand for (1+x)^n ≥ 1+nx

P[0] then stands for (1+x)^0 ≥ 1+0x. Which is true since it is equivalent to 1 ≥ 1.

Note that we only needed x ≠ -1 here because 0^0 is undefined.

So suppose P[k] is true. We will show that P[k+1] must then be true also.

Since we are given P[k] is true, then we know that (1+x)^k ≥ 1+kx.

Since x > -1, then 1 + x > 0.

So multiplying both sides of (1+x)^k ≥ 1+kx by (1 + x) does not change the

direction of the inequality. We get

(1+x)^k * (1 + x) ≥ (1+kx) (1 + x)

(1+x)^k * (1 + x)^1 ≥ 1 + kx + x + kx^2

(1 + x)^(k + 1) ≥ 1 + kx + x + kx^2

Note that we are assuming k ≥ 0 and, since it is a square, x^2 ≥ 0.

So kx^2 ≥ 0. Adding 1 + kx + x to both sides, we get

1 + kx + x + kx^2 ≥ 1 + kx + x.

It follows that

(1 + x)^(k + 1) ≥ 1 + kx + x

(1 + x)^(k + 1) ≥ 1 + (k + 1)x

Hence, if P[k] is true, it follows that P[k + 1] is true.

By mathematical induction, it follows that, for all x > -1

(1+x)^n ≥ 1+nx for all natural numbers n.

to start with, it's not real for all usual numbers, considering that it's undefined for n=one million. groundwork: n=two. one million-one million/four = three/four count on real for n. Let S(n) = product(i=two...n) (one million-one million/i^two) S(n+one million) = (one million-one million/(n+one million)^two) * S(n) via induction speculation, S(n) = (n+one million)/2n (one million-one million/(n+one million)^two)(n+one million/n2) = (n^two+2n)/(n+one million)^two *(n+one million)/2n = (n^two+2n)/(2n(n+one million)) = (n+two)/two(n+one million) and the outcome is proved for all n > one million.