Generalization of Bernoulli's Inequality Is it possible to generalize Bernoulli's Inequality to $(x+y)^n \geq x + ny$ provided $x+y \geq 0 $ and $x \geq 1$ and $n$ is a positive natural number?
I was thinking that the proof follows by induction:  In base case, have $n=1$.  Then we have that $(x+y)^n = (x+y)^1 = x+y \geq x+y$.  Now, inductively assume that $(x+y)^k \geq x + ky$.  Then $(x+y)^{k+1} = (x+y)(x+y)^k \geq (x+y)(x+ky)$ by our inductive hypothesis and as $(x+y) \geq 0$.  Now, "foiling,' we see that $(x+y)(x+ky) = x^2 + kxy + xy + ky^2 \geq x^2 + (k+1)xy$.  As $x \geq 1$, we have that $1/x \leq 1$, so $x^2 + (k+1)xy \geq (1/x)(x^2+(k+1)xy) = x + (k+1)y$.  Then by transitivity of order, we have  $(x+y)^{k+1} \geq x + (k+1)y$, closing induction.
Thus, $(x+y)^n \geq x + ny$ for all positive natural n.
Is this valid?  Did I do anything that was incorrect/invalid?  I am just curious because I have not found this generalization anywhere, and am wondering if somehow it is not a generalization at all.

There is one step that is not (always) valid. The inequality
$$x^2 + (k+1)xy \geqslant \frac{1}{x}(x^2+(k+1)xy)$$
only holds for $x+(k+1)y \geqslant 0$, if $x > 1$. If $x+(k+1)y < 0$ and $x > 1$, you have a strict inequality in the other direction,
$$x^2 + (k+1)xy < \frac{1}{x}(x^2+(k+1)xy).$$
However, in that case, the desired inequality follows from
$$(x+y)^{k+1} \geqslant 0 > x + (k+1)y.$$
For $x+ky \geqslant 0$ (for $x+ky < 0$, we have the sharper $(x+y)^k \geqslant 0 > x+ky$), the Bernoulli inequality yields a sharper bound per
$$(x+y)^k = x^k\left(1+\frac{x}{y}\right)^k \geqslant x^k\left(1+k\frac{x}{y}\right) = x^k + kx^{k-1}y = x^{k-1}(x+ky),$$
which may explain why your inequality is not mentioned in the literature.
