Which number is bigger: $e^{\sqrt{5}} + e^{\sqrt{7}}$ or $2e^{\sqrt{6}}$? Let $f:(0, + \infty) \rightarrow \mathbb{R}$ be $f(x) = e^{\sqrt{x}}$.
Which number is bigger:  $e^{\sqrt{5}} + e^{\sqrt{7}}$ or $2e^{\sqrt{6}}$?
I would like to ask for a hint on how to approach this question, I don't have an idea.
 A: Let $f(x)=e^{\sqrt{x}}$, so the second derivative is:
$$f''(x)=\frac{e^{\sqrt{x}}}{4x^{3/2}}(\sqrt{x}-1)$$
Since $f''(x)>0$ for $x\in (1,\infty)$, it is concave up (convex). So we have
$$\frac{f(x)+f(y)}{2}>f\left(\frac{x+y}{2}\right)$$
Let $x=5, y=7$
$$\frac{f(5)+f(7)}{2}>f\left(6\right)$$
A: We have
\begin{align*}
e^{\sqrt{5}} > 2^{\sqrt{5}} > 2^1 \implies \frac{1}{e^\sqrt{5}} < \frac{1}{2} \tag{$*$}\\
e^{\sqrt{7}} > 2^{\sqrt{7}} > 2^1 \implies \frac{1}{e^\sqrt{7}} < \frac{1}{2} \tag{$**$}
\end{align*}
Now because $\ln{x}$ is an increasing function, we can write:
\begin{align*}
\ln(e^{\sqrt{5}} + e^{\sqrt{7}}) &= \ln\left(\frac{e^{\sqrt{5}} + e^{\sqrt{7}}}{e^{\sqrt{5}}e^{\sqrt{7}}}\right) + \ln(e^{\sqrt{5}}) + \ln(e^{\sqrt{7}})\\
&= \ln\left(\frac{1}{e^{\sqrt{5}}} + \frac{1}{e^\sqrt{7}}\right) + \sqrt{5} + \sqrt{7}\\
(*), (**) &< \ln\left(\frac{1}{2} + \frac{1}{2}\right) + \sqrt{5} + \sqrt{7}\\
&= \sqrt{5} + \sqrt{7}\\
(?) &< 2\sqrt{6}\\
&= \ln(e^{2\sqrt{6}})
\end{align*}
Notice that for $(?)$ we can write:
\begin{align*}
\sqrt{5} + \sqrt{7} < 2\sqrt{6} &\iff (\sqrt{5} + \sqrt{7})^2 < (2\sqrt{6})^2\\
&\iff 12 + 2\sqrt{35} < 24\\
&\iff \sqrt{35} < 6
\end{align*}
