Proving $\exp(-A-B-C) \geq 1- \left[\,1 - \exp(-A) + 1-\exp(-B) + 1-\exp(-C)\,\right]$ for positive $A$, $B$, $C$

Question

Is read in a paper that the following inequality is true for any positive values $A$, $B$, and $C$. How to prove it?

$$\exp(-A-B-C) \geq 1- \left[\,1 - \exp(-A) + 1-\exp(-B) + 1-\exp(-C)\,\right]$$

Answer 1

Define $f(x) = 1-\exp(-x)$. Note that $f(0)=0$ and $f''(x) = -\exp(-x) \le 0$, so $f$ is concave.

It is well known that this implies that $f$ is subadditive, i.e. for $x, y \ge 0$, $f(x+y)\le f(x)+f(y)$.

Hence, for $A,B,C\ge 0$, $f(A+B+C) \le f(A)+f(B)+f(C)$, which is your inequality.

Answer 2

This question can be treated in a different way, by somehow "making the exponential disappear". Here is how.

Setting

$$a=e^{-A}, \ \ b=e^{-B}, \ \ c=e^{-C},$$

the inequality to be proven is

$$\forall a,b,c \in [0,1], \ \ abc \ge a+b+c-2$$

But this is a consequence of the following determinant inequality:

$$\forall a,b,c \in [0,1], \ \ \begin{vmatrix}a&1&1\\1&b&1\\1&1&c\end{vmatrix} \ge 0 \tag{1}$$

to which we give a graphical proof: indeed this determinant is equal to 6 times the volume of the positively oriented tetrahedron $OA_1B_1C_1$ (with points $A_1,B_1,C_1$ having the columns of the determinant as their coordinates) represented in the following figure:

Remarks:

1. Inequality (1) becomes an equality if and only if $a=b=c=1$.

2. (1) is extensible to any dimension.