Inequality and convex combination

Question

Suppose that $\lambda_{i}\geq 0$ and $\sum_{i=1}^{n+1}{\lambda_{i}}=1$. Does it hold that $$|\sum_{i=1}^{n+1}{\lambda_{i}x_{i}}|+\sum_{i=1}^{n}{\lambda_{i}|x_{i}|}\geq \lambda_{n+1}|x_{n+1}|$$?

Answer 1

\begin{align*} \hspace{-75pt}\Big|\sum_{i=1}^{n+1}{\lambda_{i}x_{i}}\Big|+\sum_{i=1}^{n}{\lambda_{i}|x_{i}|} &\geq |\lambda_{n+1}x_{n+1} |-\Big|\sum_{i=1}^{n}{\lambda_{i}x_{i}}\Big|+\sum_{i=1}^{n}{\lambda_{i}|x_{i}|}\tag{since $|a+b|\geq |a|-|b|$ }\\ & =|\lambda_{n+1}x_{n+1} |+\underbrace{\sum_{i=1}^{n}{\lambda_{i}|x_{i}|}-\sum_{i=1}^{n}{\lambda_{i}|x_{i}}}_{\geq 0}| \\ &\geq |\lambda_{n+1}x_{n+1} | = \lambda_{n+1}|x_{n+1} | \end{align*}

Notice that the convex sum is not needed...