Which is bigger $e^{(a+b)}$ vs $e^a + e^b$?

Question

I understand that exponential function is a convex function so for any convex function $\theta(a+b) > \theta(a) + \theta(b)$, but can someone provide a more formal proofs ?

Answer 1

Since $$ \operatorname e^{a+b} =\operatorname e^a\cdot\operatorname e^b $$ we might as well identify $x=\operatorname e^a$ and $y=\operatorname e^b$ and ask which is bigger $x\cdot y$ or $x+y$. Solving the inequality $x\cdot y<x+y$ for $y$ yields $$ y<\frac{x}{x-1}\quad\text{if}\quad x>1 $$ and $$ y>\frac{x}{x-1}\quad\text{if}\quad x<1 $$ and it always holds for $x=1$. Graphing this we get three regions in the $xy$-plane:

where the blue region corresponds to $x\cdot y<x+y$, the black curves $x\cdot y=x+y$, and the red regions $x\cdot y>x+y$. The dashed lines are the asymptotics $x=1$ and $y=1$.

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Now since $x=\operatorname e^a,y=\operatorname e^b>0$ we are only interested in the first quadrant. Transforming via logarithms from $(x,y)$ to $(a,b)=(\ln x,\ln y)$ we obtain the relationship $$ y=\frac x{x-1}\quad\iff\quad b=a-\ln(\operatorname e^a-1) $$ and since the logarithm is a strictly increasing function the inequalities are maintained so that we have: $$ \operatorname e^{a+b}>\operatorname e^a+\operatorname e^b \quad\iff\quad b>a-\ln(\operatorname e^a-1) $$ so we get the following regions of the $ab$-plane:

so when $(a,b)$ is within the red region above the curve we have $\operatorname e^{a+b}>\operatorname e^a+\operatorname e^b$, when $(a,b)$ lies on the curve we have equality of the two, and in the remainder of the $ab$-plane $\operatorname e^{a+b}<\operatorname e^a+\operatorname e^b$.

Answer 2

Fix $b \neq 0$, and consider $f(x) = e^b(1 - e^{-x})$. We notice that $e^{a + b} > e^a + e^b$ if and only if $f(a) > 1$. Since $\dfrac{df}{dx} = e^{b-x} > 0$, we find $f(a) > 1$ whenever $a > c$ where $f(c) = 1$. Simple algebra provides $c = b - \ln(e^b -1)$ which exists if $b > 0$. Suppose $b < 0$. Then $1 - e^{-b} < 0 < e^{-c}$ for every $c$ which implies $f(c) < 1$. Hence, for $a < c$, we find $f(a) < f(c) < 1$ such that $e^{a+b} < e^a + e^b$.

The take-away:

Suppose $b > 0$. If $a > b - \ln(e^b - 1)$, then $e^{a + b} > e^a + e^b$; otherwise $e^{a + b} \leq e^a + e^b$.

Suppose $b \leq 0$. The inequality $e^{a + b} < e^a + e^b$ always holds.

Here's an interesting remark: notice the symmetry between $a$ and $b$, that they are interchangeable without altering the inequality. Suppose $a$ and $b$ are both positive and satisfy the inequality $e^{a + b} > e^a + e^b$. Then $a > b - \ln(e^b - 1)$, yet this may be algebraically solved for $b > a - \ln(e^a - 1)$ .. as we would expect from interchanging $a$ and $b$.