<s>[INST] let $x > -1$ prove for all $n \geq 1$, $(1+x)^n \geq 1+nx$ I tried to prove this by induction:
Base case for n = 1 satisfies $\because$ $1+x$ = $1+x$
I.H: $(1+x)^k \geq 1 + kx$
Inductive Step for $k+1$:
$(1+x)^{k+1} \geq 1 + (k+1)x$
$(1+x)^k (1+x) \geq 1 + kx + x$
$(1+kx) (1+x) \geq 1 + kx + x$ by I.H.
$kx^2 + kx + x + 1 \geq 1 + kx + x$
$kx^2 \geq 0$ which is true for $x>-1$ therefore proved?
Is this the correct way to prove by induction?
 [/INST] You can save your proof  by added "is implied by" ($\Leftarrow$) marks between lines to indicate the desired direction:
Inductive Step for k+1
:
$(1+x)^{k+1}≥1+(k+1)x\Leftarrow$
$(1+x)^k(1+x)≥1+kx+x\Leftarrow$ by I.H.
$(1+kx)(1+x)≥1+kx+x\Leftarrow$
$kx^2+kx+x+1≥1+kx+x\Leftarrow$
$kx^2≥0$
which is true as $k \ge 1 > 0$.
This would be a correct induction step.
But my advice on a stylish and better induction step would be:
......
We are assuming that $(1+x)^k \ge 1 + kx$ and wish to show this implies $(1+x)^{k+1} \ge 1 + (k+1)x$.
$(1+x)^k \ge 1+kx$ (we are presuming this) and as $x > -1$ we know $1+x > 0$ and thus:
$(1+x)^k(1+x) \ge (1+kx)(1+x)$ so
$(1+ x)^{1+x} \ge 1 + kx + x + kx^2= 1+(k+1)x + kx^2$
And as $k$ and $x^2$ are non-negative, $kx^2 \ge 0$, so
$(1+x)^{1+x} \ge 1+(k+1)x + kx^2 \ge 1+(k+1)x$.
......
but ultimately the proof will need to be written by you.
......
I just realized no-one has yet addressed that claiming $(1+x)^k(1+x) \ge (1+kx)(1+x)$ requires knowing that $1+x > 0$.  This is why the condition $x > -1$ is required.
</s>