Let's say I have a random variable $X$ with $3$ outcomes $x_1, x_2, x_3$ with the following probabilities:
$$ p(x_1) = .2 \\ p(x_2) = .3 \\ p(x_3) = .5. $$
Now I want to prove that a convex combination of each of these probabilities is at most the largest value $.5$, meaning, I want to prove the following:
$$\lambda_1 \times .2 + \lambda_2 \times .3 + \lambda_3\times .5 \le .5$$, where $\lambda_i \in [0, 1]$ and $\sum_i \lambda_i = 1$. One way I have tried is the following. I could simply drop the minimum value $.2$ and renormalize the rest like the following:
$$ \frac{\lambda_2}{\lambda_2 + \lambda_3} \times .3 + \frac{\lambda_3}{\lambda_2 + \lambda_3} \times .5, $$ which again is less than $.5$. This does not seem like a principled approach to go though. What would be a better way? Thanks in advance.
\begin{align} \sum_{i=1}^k \lambda_k p_k \le \sum_{i=1}^k \lambda_k p_{max}= p_{max}\left(\sum_{i=1}^k \lambda_k \right)=p_{\max}(1) = p_{\max} \end{align}
where we have use non-negativity of $\lambda_k$ in the first inequality.