Prove inequality - Convex problem

Question

Let $\{q_i\}_{i=1}^n$ be a sequence such that $0\le q_1\le q_2\le \cdots\le q_n$. How can I prove that $$\sum_{i=1}^n q_is_i\le q_n,$$ where $\sum\limits_{i=1}^n s_i = 1$ for all $s_i\ge 0$?

I tried Induction @Taroccoesbrocco – Alex Jun 23 '18 at 17:16 — Alex, Jun 23 '18 at 17:16
I accept Paprika answer. It is very simple. – Alex Jun 24 '18 at 06:55 — Alex, Jun 24 '18 at 06:55

score 0 · Accepted Answer · answered Jun 23 '18 at 16:55

0

The sequence satisfies $q_i\leq q_n\;\forall\, i\in\{1,\,\ldots,\,n\}$.

Now consider these manipulations: $$\sum_{i=1}^n q_i s_i\leq\sum_{i=1}^n q_n s_i=q_n\sum_{i=1}^n s_i=q_n.$$

answered Jun 23 '18 at 16:55

Paprika

45

Prove inequality - Convex problem

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