Proof for which exponent is greater

Question

Is there a way to prove which one of these is bigger? $e^{(a+b)}$ or $e^a + e^b$?

Thanks

Answer 1

$$\frac{(e^a + e^b)}{2} \geq e^{\frac{(a+b)}{2}}\text{ (Using A.M.-G.M. inequality.)}$$ $$(e^a + e^b) > e^{\frac{(a+b+1)}{2}}\text{ (Using $4>e$).}$$ If, $e^{\frac{(a+b+1)}{2}}\leq 1$, then $$e^{(a+b+1)}\leq e^{\frac{(a+b+1)}{2}} \text{ (Using the fact that, if $x \in [0,1]$, then } x\geq x^2).$$ So, $$(e^a + e^b)> e^{(a+b+1)} \text{ when, }a+b+1 \leq 0.$$ $$\implies (e^a + e^b)> e^{(a+b)} \text{ when, }a+b+1 \leq 0.$$

Answer 2

$$xy>x+y\iff x(y-1)>y$$

If $x,y>0,$

If $y-1>0, x(y-1)>y\iff x>\dfrac y{y-1}$

Else $x(y-1)>y\iff x<\dfrac y{y-1}$

Answer 3

Dividing through by $e^{a+b}$ gives $$ e^a+e^b \lessgtr e^{a+b} \qquad\text{as}\qquad e^{-a}+e^{-b} \lessgtr 1$$ This is not much nicer, but it is somewhat nicer because there's now only one $a$ and one $b$. So it is now easier to solve for $a$ or $b$: $$ e^a+e^b \lessgtr e^{a+b} \qquad\text{as}\qquad a \gtrless \begin{cases} -\log(1-e^{-b}) &\text{when } b> 0 \\ \infty & \text{when }b\le 0 \end{cases}$$