Proving an inequality regarding increasing or decreasing sequences

Question

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?

Answer 1

$$ a_2 b_1 + a_1 b_2 \leq a_1 b_1 + a_2 b_2 \iff 0 \leq (a_1-a_2) b_1 + (b_1 - b_2) a_1$$ is wrong.

What you get is $$ a_2 b_1 + a_1 b_2 \leq a_1 b_1 + a_2 b_2$$ $$ \iff 0 \leq (a_1-a_2) b_1 + (a_2 - a_1) b_2 \iff 0 \leq (a_1-a_2) (b_1-b_2)$$ which is true.

Answer 2

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.