TITLE: Proving an inequality regarding increasing or decreasing sequences
QUESTION [1 upvotes]: Suppose $a_1 \geq a_2 $ and $b_1 \geq b_2$. Prove that
$$ (a_1+a_2)(b_1+b_2) \leq 2 (a_1b_1 + a_2 b_2) $$
Generalize

sol attempt:
We can write $a_1 b_1 + a_2 b_2 + a_2 b_1 + a_1 b_2 \leq 2a_1 b_1 + 2 a_2 b_2 $
and so
$$ a_2 b_1 + a_1 b_2 \leq a_1 b_1 + a_2 b_2 \iff  0 \leq (a_1-a_2) b_1 + (b_1 - b_2) a_1$$
Now, certainly $a_2 - a_1 \geq 0$ and $b_1 - b_2 \geq 0$ by hypothesis. So, as long as $b_1 \geq 0$ and $a_1 \geq 0$ we are done. If by contradiction $b_1 < 0$ then $b_1 < b_1 - b_2 $ and so $b_2 < 0$ but I dont see how to get a contradiction from here...
I guess we have to consider cases?

REPLY [1 votes]: It's a private case of the Chebyshov's inequality:

Let $a_1\geq a_2\geq ...\geq a_n$ and $b_1\geq b_2\geq...\geq b_n$. Prove that:
  $$n(a_1b_1+a_2b_2+...+a_nb_n)\geq(a_1+a_2+...+a_n)(b_1+b_2+...+b_n).$$

For $n=2$ we obtain your inequality.
A proof of the Chebyshov's inequality we get immediately by Rearrangement, which we can prove by induction.
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