First of all, it's clear that defining $$ \lambda _{k}=\frac{\mu _{k}}{\sum_{k=1}^{n}\mu _{k}} $$ we have $$ \sum_{k=1}^{n}\lambda _{k}=1 $$

so it will we enough to prove only the simplified version.

Let's proceed by induction.

1) $n=2$ we have to show that, for any $x_{1}$ $x_{2}$ in $[a,b]$ $$ f\left( \lambda _{1}x_{1}+\lambda _{2}x_{2}\right) \leq \lambda _{1}f(x_{1})+\lambda _{2}f(x_{2}). $$ But, since $\lambda _{1}+\lambda _{2}$ must be equal to 1, we can put $% \lambda _{2}=1-\lambda _{1}$ so that the thesis becomes $$ f\left( \lambda _{1}x_{1}+\left( 1-\lambda _{1}\right) x_{2}\right) \leq \lambda _{1}f(x_{1})+\left( 1-\lambda _{1}\right) f(x_{2}), $$ which is true by definition of a convex function.

2) Taking as true that $f\left( \sum_{k=1}^{n-1}\mu _{k}x_{k}\right) \leq \sum_{k=1}^{n-1}\mu _{k}f\left( x_{k}\right) $ where $\sum_{k=1}^{n-1}\mu _{k}=1$ we have to prove that $$ f\left( \sum_{k=1}^{n}\lambda _{k}x_{k}\right) \leq \sum_{k=1}^{n}\lambda _{k}f\left( x_{k}\right) , $$ where $\sum_{k=1}^{n}\lambda _{k}=1$

First of all, let's observe that $$ \sum_{k=1}^{n-1}\frac{\lambda _{k}}{1-\lambda _{n}}=\frac{\left( \sum_{k=1}^{n}\lambda _{k}\right) -\lambda _{n}}{1-\lambda _{n}}=\frac{% 1-\lambda _{n}}{1-\lambda _{n}}=1 $$ and that if all $x_{k}\in \lbrack a,b]$ $\sum_{k=1}^{n-1}\frac{\lambda _{k}% }{1-\lambda _{n}}x_{k}$ belongs to $[a,b]$ as well. In fact, $\frac{\lambda _{k}}{1-\lambda _{n}}$ being non-negative, $$ a\leq x_{k}\leq b\Rightarrow \frac{\lambda _{k}}{1-\lambda _{n}}a\leq \frac{% \lambda _{k}}{1-\lambda _{n}}x_{k}\leq \frac{\lambda _{k}}{1-\lambda _{n}}b, $$ and, summing over $k$ $$ a\sum_{k=1}^{n-1}\frac{\lambda _{k}}{1-\lambda _{n}}\leq \sum_{k=1}^{n-1}% \frac{\lambda _{k}}{1-\lambda _{n}}x_{k}\leq b\sum_{k=1}^{n-1}\frac{\lambda _{k}}{1-\lambda _{n}}, $$ that is$$ a\leq \sum_{k=1}^{n-1}\frac{\lambda _{k}}{1-\lambda _{n}}x_{k}\leq b. $$

We have, by definition of a convex function: \begin{eqnarray*} f\left( \sum_{k=1}^{n}\lambda _{k}x_{k}\right) &=&f\left( \sum_{k=1}^{n-1}\lambda _{k}x_{k}+\lambda _{n}x_{n}\right) \\ &=&f\left( \left( 1-\lambda _{n}\right) \sum_{k=1}^{n-1}\frac{\lambda _{k}}{% 1-\lambda _{n}}x_{k}+\lambda _{n}x_{n}\right) \\ &\leq &\left( 1-\lambda _{n}\right) f\left( \sum_{k=1}^{n-1}\frac{\lambda _{k}}{1-\lambda _{n}}x_{k}\right) +\lambda _{n}f\left( x_{n}\right) . \end{eqnarray*}But, by inductive hypothesis, since $\sum_{k=1}^{n-1}\frac{\lambda _{k}}{% 1-\lambda _{n}}=1$ we have: $$ f\left( \sum_{k=1}^{n-1}\frac{\lambda _{k}}{1-\lambda _{n}}x_{k}\right) \leq \sum_{k=1}^{n-1}\frac{\lambda _{k}}{1-\lambda _{n}}f\left( x_{k}\right) , $$ so that\begin{eqnarray*} f\left( \sum_{k=1}^{n}\lambda _{k}x_{k}\right) &\leq &\left( 1-\lambda _{n}\right) f\left( \sum_{k=1}^{n-1}\frac{\lambda _{k}}{1-\lambda _{n}}% x_{k}\right) +\lambda _{n}f\left( x_{n}\right) \\ &\leq &\left( 1-\lambda _{n}\right) \sum_{k=1}^{n-1}\frac{\lambda _{k}}{% 1-\lambda _{n}}f\left( x_{k}\right) +\lambda _{n}f\left( x_{n}\right) \\ &=&\sum_{k=1}^{n-1}\lambda _{k}f\left( x_{k}\right) +\lambda _{n}f\left( x_{n}\right) \\ &=&\sum_{k=1}^{n}\lambda _{k}f\left( x_{k}\right) \end{eqnarray*}which is the thesis.