For positive reals $a_i$ and $b_i$, if $\sum a_i \geq \sum a_i b_i$, then $\sum a_i \leq \sum\frac{a_i}{b_i}$

Question

Given $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$ and $b_1$, $b_2$, $b_3$, $\ldots$, $b_n$ positive real numbers such that $$a_1 + a_2 + a_3 +\cdots+ a_n \geq a_1b_1+a_2b_2+a_3b_3...a_nb_n$$ show that: $$a_1 + a_2 + a_3+\cdots+a_n \leq \frac{a_1}{b_1} + \frac{a_2}{b_2} + \frac{a_3}{b_3}+\cdots+\frac{a_n}{b_n}$$

This was an admissions question for a math camp I ended up going to. After the program, I'm still wondering how to prove this. (The hint I got at the camp was to use Cauchy-Schwarz, but not sure where.)

I would love hearing solutions both with and without C-S, hope you can help!

Answer 1

By C-S $$\sum_{k=1}^n\frac{a_k}{b_k}\sum_{k=1}^na_kb_k\geq\left(\sum_{k=1}^na_k\right)^2\geq\sum_{k=1}^na_k\sum_{k=1}^na_kb_k$$ and we are done!

Answer 2

I think this will work. I'm concerned that I didn't apply the CS inequality and therefore made a mistake near the last step.

As discussed in the comments

$$a_1 + a_2 + a_3...a_n \geq {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}...{a_n}{b_n}$$

can be rewritten as

$$a_1\Big(\frac{1-b_1}{b_1}\Big)+a_2\Big(\frac{1-b_2}{b_2}\Big)+\dots+a_n\Big(\frac{1-b_n}{b_n}\Big)\geq 0$$

then as $a_i,b_i (i=1,2,\dots n)$ are positive real numbers, we need to consider what happens to

$$a_i\Big(\frac{1-b_i}{b_i}\Big)$$

by analyzing three different cases.

Case 1: $~0 < a_i,b_i < 1$. In this case,

$$a_i\Big(\frac{1-b_i}{b_i}\Big) >a_i\Big({1-b_i}\Big)$$

since $0 < b_i < 1$.

Case 2: $a_i=b_i=1$. In this case,

$$a_i\Big(\frac{1-b_i}{b_i}\Big) =a_i\Big({1-b_i}\Big)=0$$

because $b_i = 1$.

Case 3: $~a_i,b_i > 1$. In this case,

$$a_i\Big(\frac{1-b_i}{b_i}\Big) >a_i\Big({1-b_i}\Big)$$

because both are negative but the LHS will be a smaller negative number.

This allows us to conclude that

$$\sum_{i=1}^n a_i\Big(\frac{1-b_i}{b_i}\Big) > \sum_{i=1}^n a_i\Big({1-b_i}\Big)$$

which can be rearranged to (provided $b_i \neq 1$ in which case the sum is $0$)

$$\sum_{i=1}^n \bigg(\frac{a_i\Big(\frac{1-b_i}{b_i}\Big)}{1-b_i}\bigg) > \sum_{i=1}^n a_i$$

so that

$$\sum_{i=1}^n a_i < \sum_{i=1}^n \Big(\frac{a_i}{b_i}\Big)$$

Answer 3

(I consider this different from Michael's solution, but let me know if that's not the cse.)

Proof by contradiction.

Suppose not, so $ \sum a_i > \sum \frac{a_i}{b_i}$.
Then we have the following contradiction:

$$ \sum a_i b_i \sum \frac{a_i}{b_i} \geq (\sum a_i)^2 > \sum a_i b_i \sum \frac{a_i }{b_i }$$