How do you prove by induction that for any real number x>= -1 and any natural number n, (1+x)^n >= 1 + nx?

Proof by induction always consists of two steps:
1) Prove that the equation (or in this case, inequality) is true for start values of n
2) Prove that if the equation (or inequality) is true for n, then the equation (/inequality) is true for n+1.

Step 1: inequality is true for start value of n.
Since n is limited to natural numbers, the start value of n is n = 0.

For n = 0, the inequality (1+x)^n ≥ 1 + nx becomes

(1+x)^0 ≥ 1 + 0*x
1 ≥ 1

which is true, hence the inequality is true for n = 0.

Step 2: validity of inequality for n implies validity of inequality for n+1:
Or in other words: for x ≥ -1 and n ≥ 0, given that (1+x)^n ≥ 1 + nx, prove that also

(1+x)^(n+1) ≥ 1 + (n+1)x

We'll start with the induction hypothesis (the inequality that we're assuming to be true):

(1+x)^n ≥ 1 + nx

Since x ≥ -1, it means (1 + x) is non-negative, hence the ≥ sign won't flip when we multiply both sides by (1+x).

(1+x)^n * (1+x) ≥ (1 + nx)*(1+x)

LHS equals (1+x)^(n+1). RHS equals 1 + nx + x + n*x^2 = 1 + (n+1)x + n*x^2 .

(1+x)^(n+1) ≥ 1 + (n+1)x + n*x^2

The term n*x^2 in the RHS is non-negative (because x^2 is a square so x^2 ≥ 0, and n was assumed to be a natural number), so RHS is greater than or equal to 1 + (n+1)x.

(1+x)^(n+1) ≥ 1 + (n+1)x + n*x^2 ≥ 1 + (n+1)x

So we have shown that (1+x)^n ≥ 1 + nx also implies (1+x)^(n+1) ≥ 1 + (n+1)x, which is what we were required to prove.

We have now completed step 1 and step 2 of the proof by induction, hence we've proven that the inequality is true for all natural n.

You may already know, to prove by induction, we have to follow 3 steps.
1st, prove it is true for n=1
2nd, assume it is true for n=k
3rd, prove it is true for n=k+1 if the 2nd step is true.

So,
1st, substitute n=1 to the equation:
(1+x)^n >= 1 + nx
=> (1+x)^1>=1+(1)x
=> 1+x>=1+x is true
2nd, assume (1+x)^k >= 1 + kx is true
3rd, substitute k+1 to LHS of the equation:
(1+x)^n
=> (1+x)^(k+1)
=> (1+x)^k * (1+x) ........... look back to 2nd step, (1+x)^k >=1+kx
so, (1+x)^(k+1) >=(1+kx)(1+x)
(1+x)^(k+1)>=1+kx+x+kx^2
(1+x)^(k+1)>=1+(k+1)x+kx^2
(1+x)^(k+1)>=1+(k+1)x
Thus, by the principle of mathematical induction, the equation is true for all natual numbers n.

suppose it's true for n = k , ie, (1+x)^k >= 1 + kx
so (1+x)^(k+1) >= (1 + kx)(1+x) = 1+kx^2 + (k+1)x >= 1 + (k+1)x ---> it's true for n=k+1

hence for any real number x>= -1 and any natural number n, (1+x)^n >= 1 + nx