Adding increasing fractions without averaging numerators

[Afterthought]

I'm interested in the following inequality (which may or may not be true)

Theorem 1:
$( \sum_{i=1}^n \frac{a_i} {n}\ )( \sum_{i=1}^n \frac{1} {b_i}\ ) > \sum_{i=1}^n \frac{a_i} {b_i}\$
Where $n ≥ 2, a_1 < a_2 < ... < a_n$ and $b_1 < b_2 < ... < b_n$.

My attempt at a proof:

1) When n = 2, this says that $(\frac{a_1 + a_2} {2})(\frac{1} {b_1}\ + \frac{1} {b_2}) > \frac{a_1} {b_1} + \frac{a_2} {b_2}$
2) Expanding gives $\frac{a_1}{2b_1} + \frac{a_2}{2b_1} + \frac{a_1}{2b_2} + \frac{a_2}{2b_2} > \frac{a_1} {b_1} + \frac{a_2} {b_2}$
3) Because $a_2 > a_1$, we can replace $a_2$ with $a_1 + x$, where x is a positive real number, and get $\frac{a_1}{2b_1} + (\frac{a_1}{2b_1} + \frac{x}{2b_1}) + (\frac{a_2}{2b_2} - \frac{x}{2b_2}) + \frac{a_2}{2b_2} > \frac{a_1} {b_1} + \frac{a_2} {b_2}$
4) Combining terms and rearranging, we get $\frac{a_1} {b_1} + \frac{a_2} {b_2} + \frac{x}{2b_1} - \frac{x}{2b_2} > \frac{a_1} {b_1} + \frac{a_2} {b_2}$
5) Since $b_2 > b_1, \frac{x}{2b_1} - \frac{x}{2b_2} > 0$, and then the above equality is true.

With a base case of n = 2 proven, I will now use induction to generalize this result:

6) Assume the inequality is true for n-1: $( \sum_{i=1}^{n-1} \frac{a_i} {n-1}\ )( \sum_{i=1}^{n-1} \frac{1} {b_i}\ ) > \sum_{i=1}^{n-1} \frac{a_i} {b_i}\$
7) Setting z = $\sum_{i=1}^{n-1} a_i\$, we can rewrite this as $( \sum_{i=1}^{n-1} \frac{z} {(n-1)b_i}\ ) > \sum_{i=1}^{n-1} \frac{a_i} {b_i}\$
8) Adding $\frac{a_n} {b_n}$ to both sides gives $( \sum_{i=1}^{n-1} \frac{z} {(n-1)b_i}\ ) + \frac{a_n} {b_n} > \sum_{i=1}^{n} \frac{a_i} {b_i}\$
9) I will now make the following assumption: $( \sum_{i=1}^{n} \frac{z + a_n} {nb_i}\ ) > ( \sum_{i=1}^{n-1} \frac{z} {(n-1)b_i}\ ) + \frac{a_n} {b_n} > \sum_{i=1}^{n} \frac{a_i} {b_i}\$
By showing that the first term is greater than the second term, we can show that it is also greater than the third, and then we will be done (as the first term is another way of writing the first term in Theorem 1).

The rest is just algebra:

10) Changing denominators: $( \sum_{i=1}^{n} \frac{(n-1)(z + a_n)} {n(n-1)b_i}\ ) > ( \sum_{i=1}^{n-1} \frac{nz} {n(n-1)b_i}\ ) + \frac{n(n-1)a_n} {n(n-1)b_n}$
11) Expanding the first term: $( \sum_{i=1}^{n-1} \frac{nz + na_n - z - a_n} {n(n-1)b_i}\ ) + \frac{(n-1)(z + a_n)} {n(n-1)b_n} > ( \sum_{i=1}^{n-1} \frac{nz} {n(n-1)b_i}\ ) + \frac{n(n-1)a_n} {n(n-1)b_n}$
12) Rearranging: $( \sum_{i=1}^{n-1} \frac{nz + na_n - z - a_n} {n(n-1)b_i}\ ) - ( \sum_{i=1}^{n-1} \frac{nz} {n(n-1)b_i}\ ) > \frac{n(n-1)a_n} {n(n-1)b_n} - \frac{(n-1)(z + a_n)} {n(n-1)b_n}$
13) The above becomes $( \sum_{i=1}^{n-1} \frac{na_n - z - a_n} {n(n-1)b_i}\ ) > \frac{na_n - z - a_n} {nb_n}$
14) Diving both sides by $na_n - z - a_n$: $( \sum_{i=1}^{n-1} \frac{1} {n(n-1)b_i}\ ) > \frac{1} {nb_n}$
15) Since $b_{n-1} > b_{n-2} > ... > b_1$, $( \sum_{i=1}^{n-1} \frac{1} {n(n-1)b_i}\ ) > \sum_{i=1}^{n-1} \frac{1} {n(n-1)b_{n-1}} = \frac{1} {nb_{n-1}}$. Similary, $\frac{1} {nb_{n-1}} > \frac{1} {nb_{n}}$

Therefore, the inequality in step 14 is true, which means that the inequality in step 9 is true, which proves Theorem 1.
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This was my first attempt at a formal proof, so sorry for any mistakes! My questions:

1) Is this a valid proof? Are there any missing steps or mistakes?
2) Is there a simpler proof?
3) Has this inequality been noticed before? Are there any other related stuff I can read up on?

Thanks.

 

[mfb]

I didn't check each individual expression, but the general idea is right and the key steps look right as well.
If you replace the b_i by their inverse (and require that those are in decreasing order), it is a common inequality. A nice homework problem for undergrads I guess.

It looks related to the problem "show that a/b+b/c+c/a >= 3".