= Solution
The <mean value theorem> says that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then $f(b)-f(a)=f'(c)(b-a)$ for some $c\in(a,b)$.
For $F(t)=\cos(e^{-t})$ and $t\geq0$,
$$
|F'(t)|=e^{-t}|\sin(e^{-t})|\leq1.
$$
The mean value theorem gives \b[$|\cos(e^{-x})-\cos(e^{-y})|\leq|x-y|$].
For the second inequality put $t=\sqrt{1+x}\geq1$. It becomes $2\log t\leq t-t^{-1}$. The difference $G(t)=t-t^{-1}-2\log t$ satisfies $G(1)=0$ and
$$
G'(t)=1+t^{-2}-2t^{-1}=\frac{(t-1)^2}{t^2}\geq0.
$$
Thus \b[$\log(1+x)\leq x/\sqrt{1+x}$].
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