Is it true that if $x \ge \min \{a,b\}$ and $a \ge \min \{a_1,\ldots,a_n\}$ then $x \ge \min \{a_1,\ldots,a_n,b\}$?

Question

Let $f:X \to \mathbb R$ be a quasi-concave function. It follows that $f(\lambda x+(1-\lambda) y) \geq \min \{f(x), f(y)\}$ for all $\lambda \in [0,1]$ and $x,y \in X$.

I'm trying to prove that for all $x_i \in X$ and $\lambda_i \ge 0$ such that $\sum_{i=1}^n \lambda_i = 1$, we have $$f \left (\sum_{i=1}^n \lambda_i x_i \right) \ge \min_{1 \le i \le n} f(x_i)$$

This leads me to prove that if $x \ge \min \{a,b\}$ and $a \ge \min \{a_1,\ldots,a_n\}$ then $x \ge \min \{a_1,\ldots,a_n,b\}$.

My question:

Could you please shed me some lights on how to prove the last claim? Thank you so much!

Answer 1

If on the contrary $x<\min\{a_1,\ldots, a_n,b\}$ then $x<a_1,x<a_2\ldots x<a_n,x<b$. From the first $n$ inequalities, $x<\min\{a_1,\ldots, a_n\}\le a$ and from $x<a$ and $x<b$, we find $x<\min\{a,b\}$.

Answer 2

Prove by contradiction. If the result is false then $x <b$ and $x <a_i$ for each $i$. Hence $x <\min \{a_1,a_2,...,a_n\} \leq a$. Do you see a contradiction now?

Answer 3

Well, suppose that it's false. Then $a_1,...,a_n,b$ are all strictly greater than $x$. Since $a\geq\min\{a_1,..,a_n\}$ there is some $k$ for which $a\geq a_k>x$. Hence $a$ and $b$ are both greater than $x$ which is a contradiction.