1. ## Prove

Prove that $e^x \geqslant x+1$

using the taylor series for all $x > 0$ i can use the taylor series for $e^x$ its obvious that the inequality is true for positive values.

then using the fact that $e^x > 0$ works for all x less than -1.

but i am jut a bit stuck on proving the inequality for values of x $-1 \leqslant x \leqslant 0$

2. Do not use Taylor series. Instead define $f(x) = e^x - x -1$ and show that $f'(x) > 0$ on $(0,\infty)$ now argue that it is posititive.

3. thank you TPH I see how that argument is better, but it does not prove the inequality for $-1 \leqslant x \leqslant 0$

or did i missed something ?

4. Originally Posted by bobak
thank you TPH I see how that argument is better, but it does not prove the inequality for $-1 \leqslant x \leqslant 0$

or did i missed something ?
TPH's method will work for all x. just show that the derivative is positive for x < 0 as well (try it, it's easier than you think)

note: we have equality when x = 0

5. Originally Posted by Jhevon
TPH's method will work for all x. just show that the derivative is positive for x < 0 as well (try it, it's easier than you think)
It is not quite that easy. The derivative is of $e^x - x -1$ is negative if $x<0$.
However, that function has an absolute minimum at $x=0$.
Hence that function is always non-negative.

6. Originally Posted by Plato
It is not quite that easy. The derivative is of $e^x - x -1$ is negative if $x<0$.
However, that function has an absolute minimum at $x=0$.
Hence that function is always non-negative.
indeed. i had graphed the function and everything and noticed it was non-negative. mixed that up with the derivative being positive somehow.

7. Its all very clear now, thanks a lot.

8. Read G. Pólya's proof of the Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia
it is interesting and involves the inequality $e^x\geq{1+x}$