How to prove that 1-x <= e^-x?

We will prove that:

1 - x ≤ e^-x; -∞ < x < ∞

To do this, we will "break up" the domain, -∞ < x < ∞, into within 3 domain segments and a single point as follows:

1 - x ≤ e^-x; x ≤ 0, 0 < x < 1, x = 1, and x > 1


Let's begin by examining:

1 - x ≤ e^-x; x ≤ 0

This is the same as:

1 + x ≤ e^x; x ≥ 0 (see footnote)

Substitute e^x = 1 + x + (x^2)/(2!) +( x^3)/(3!) + ...


1 + x ≤ 1 + x + (x^2)/(2!) +(x^3)/(3!) + ...

subtract 1 + x from both sides:

0 ≤ (x^2)/(2!) + (x^3)/(3!) + ... ; x ≥ 0

This is certainly true within the specified domain segment.

Now let's examine:

1 - x ≤ e^-x; 0 < x < 1

substitute e^-x = 1 - x + (x^2)/(2!) - (x^3)/(3!) + (x^4)/(4!) - ...

1 - x ≤ 1 - x + (x^2)/(2!) - (x^3)/(3!) + (x^4)/(4!) - ...; 0 < x < 1

subtract 1 - x from both sides:

0 ≤ (x^2)/(2!) - (x^3)/(3!) + (x^4)/(4!) - ...; 0 < x < 1

the above series can be written as the difference of two summations:

∞
Σ (x^{2n})/({2n}!)
k=1

minus

∞
Σ (x^{2n+1})/({2n+1}!)
k=1

We observe that x^{2n+1} is always less than x^{2n}, when 0 < x < 1, and {2n+1}! is always greater than {2n}!.

Therefore, the difference between these two series must be a positive number; this makes the inequality true in this domain segment.

Now we examine:

1 - x ≤ e^-x; x = 1

1 - 1 ≤ e^-1

0 ≤ e^-1

This is certainly true.

Finally, we examine:

1 - x ≤ e^-x; x > 1

In this domain segment, 1 - x, becomes negative and e^-x approaches a 0 asymptote from the positive direction but never becomes negative. Therefore, the inequality is true for this domain segment.

By showing that:

1 - x ≤ e^-x; x ≤ 0, 0 < x < 1, x = 1, and x > 1

We have shown that,

1 - x ≤ e^-x; -∞ < x < ∞

Q.E.D.

(Footnote: At first glance, one might say that; "Because I have changed the 1 - x to 1 + x and e^-x to e^x, I have made an error or should I change the direction of the inequality". But this is not so. Because I have, also, changed the domain segment to be one-for-one equivalent but opposite in sign, this is an implicit change of variables; not an error or a multiplication by negative one.)

let x be negative. Then e^-x = 1 + |x| + x^2/2 + |x|^3/3! +...., and since x =! 0, this quantity is always greater than 1 + |x| = 1 - x.

now for x > 0.

suppose e^-x < 1-x. Then
1 - x + x^2/2 - x^3/6 + ... < 1-x
x^2/2 - x^3/6 + x^4/24 - ... < 0
1/2 - x/6 + x^2/24 -... < 0 since x =! 0
then x^2/24 - x^3/120 + ... < x/6 - 1/2

however, this is a contradiction, because for small x, the left hand side is zero while the right hand side is negative 1/2.

we also know that e^0 = 1 - 0 = 1, so e^-x >= 1-x

this isn't a very elegant proof, but it's what comes to mind. You could probably find a nicer way of doing it without relying on calculus.