Prove the generalization of Bernoulli's inequality $$
1+nx \le (1+x)^n, \forall x \ge-2, \forall n \in \Bbb N
$$  
When proving this:
$P(1): 1+n(-2) \le (1+(-2))^n =>$ Let $n = 1$, then $-1 \le -1$ which is true.
$P(n+1): ...$ This part I know how to do, but my question is if the only difference is that the -2 works just as the -1 in the for all x?
 A: My favorite way of attacking the Bernoulli inequality is via a geometric sum. The formula for geometric sum gives
$$
(1+x)^n-1=\bigl[1+(1+x)+(1+x)^2+\cdots+(1+x)^{n-1}\bigr]x.
$$ 
If $x>0$ then we have $n$ terms inside the square bracket that are all at least $1$, and hence the right hand side is $\geq nx$.
If $-2\leq x\leq0$ then $-1\leq 1+x\leq1$. Hence $-1\leq(1+x)^k\leq 1$ for all $k\geq 1$, which implies that the sum inside the brackets is bounded by $n$, i.e.
$$
\bigl[1+(1+x)+(1+x)^2+\cdots+(1+x)^{n-1}\bigr]\leq n.
$$
Multiplying this with $x$, which is non-positive, gives
$$
\bigl[1+(1+x)+(1+x)^2+\cdots+(1+x)^{n-1}\bigr]x\geq nx,
$$
and we are done.
A: You can also take this approach of proving the inequality:
Expand $$(1+x)^n= 1+nx+ \sum_{i=2}^{n}\binom{n}{i}x^i $$$$\leq 1+nx$$ for $x>0$.
Now when $-1<x<0$,
$$(1-x)^n= 1-nx+\sum_{i=2}^{n}(-1)^i\binom{n}{i}x^i$$ Here you can see that the coefficients of the subsequent terms are smaller from the earlier ones. This implies the sum will be some positive number.
So we can say that   $$(1-x)^n\geq 1-nx$$ for $-1<x<0$.
Hence we can say that $$(1+nx)\leq (1+x)^n$$
A: We can also do this (for $0\leq r\leq1$) by the weighted A-M and G-M inequality. 
Let $a_1,a_2$ be positive constants(weights in this case) on $1,1+x$ . Now using weighted AM -GM inequality:
$$\frac{a_1+(1+x)a_2}{a_1+a_2} \geq \sqrt[a_1+a_2]{(1+x)^{a_2}}$$
The left hand side after a few algebraic manipulations becomes $$1+\frac{a_2}{a_1+a_2}x$$
And by the law of exponents the right hand side becomes ${(1+x)^{\frac{a_2}{a_1+a_2}}}$. So now the inequality becomes $$1+\frac{a_2}{a_1+a_2}x \geq {(1+x)^{\frac{a_2}{a_1+a_2}}}$$.
Let $\frac{a_2}{a_1+a_2}=r$
So the inequality becomes $$1+rx\geq (1+x)^r$$
