Proof (without use of differential calculus) that $e^{\sqrt{x}}$ is convex on $[1,+\infty)$. 
Prove (without use of  differentiation) that $f(x)=e^{\sqrt{x}}$ is convex on $[1,+\infty)$.

Attempt. Function $x\mapsto e^x$ is convex and increasing, but $x\mapsto \sqrt{x}$ is concave, so we cannot use the proposition of composition: $$(convex~\&~increasing)\circ convex=convex.$$
Definition would require to prove for all $x,~y\geqslant 1$ and $\lambda \in [0,1]$: $$e^{\sqrt{\lambda x+(1-\lambda) y}}\leqslant \lambda e^{\sqrt{ x}}+(1-\lambda)e^{\sqrt{y}}$$
but squaring doesn't work here. 
If we used continuity in order to prove mid-convexity, the problem would go like: 
$$e^{\sqrt{\frac{x+y}{2}}}\leqslant \frac{e^{\sqrt{x}}+e^{\sqrt{y}}}{2},$$
equivalently:
$$e^{\sqrt{\frac{x+y}{2}}}-e^{\sqrt{x}} \leqslant  e^{\sqrt{y}}-e^{\sqrt{\frac{x+y}{2}}}$$
(but without MVT what would we do with these differences?)   
Thanks in advance for the help.
 A: To show convexity it is enough to prove the property of a supporting hyperplane at each point, i.e.
$$e^{\sqrt{x+t}}\geq e^{\sqrt x}+t\cdot \tfrac{1}{2\sqrt x}e^{\sqrt x}\tag{*}$$
whenever $x\geq 1$ and $t+x\geq 1.$
I will just recall here how your definition of convexity follows from (*), to show that it does not use calculus. Given $y,z\geq 1$ and $0\leq \lambda\leq 1,$ set $x=\lambda y+(1-\lambda)z.$ Then (*) gives
\begin{align*}
e^{\sqrt y}&\geq e^{\sqrt{x}} + (y-x)\cdot \frac{1}{2\sqrt x}e^{\sqrt x}\\
e^{\sqrt z}&\geq e^{\sqrt{x}} + (z-x)\cdot \frac{1}{2\sqrt x}e^{\sqrt x}\\
&\implies\\
\lambda e^{\sqrt y}+(1-\lambda)e^{\sqrt z}&\geq e^{\sqrt{x}} + (\lambda(y-x)+(1-\lambda)(z-x))\cdot \frac{1}{2\sqrt x}e^{\sqrt x}\\
&= e^{\sqrt{x}}.
\end{align*}

It remains to show (*).
For $t\leq 0$ we can use
$$\sqrt{x+t}-\sqrt{x}=\frac{t}{\sqrt{x+t}+\sqrt{x}}\geq \frac{t}{2\sqrt x}$$
to get $e^{\sqrt{x+t}-\sqrt x}\geq e^{t/2\sqrt x}\geq 1+t/2\sqrt x,$ which is (*).
For $t\geq 0$ we can find an $h\geq 0$ such that $1+\tfrac{t}{2\sqrt x}=e^h.$ Since $h\geq 0$ we have $e^h\geq 1+h+h^2/2,$ so
\begin{align*}
x+t&=x+2\sqrt{x}(e^h-1)\\
&\geq x+2h\sqrt x+h^2\sqrt x\\
&\geq x+2h\sqrt x+h^2\\
&=(h+\sqrt x)^2.
\end{align*}
(This is where we need $x\geq 1.$) Taking square roots and exponentiating gives $e^{\sqrt{x+t}}\geq e^{h+\sqrt x},$ which is (*).
