AoPS Academy
from the nonnegative integers to the integers satisfying 
and ![\[ f(mn) = f(m) + f(n) + f(m)f(n) \]](http://latex.artofproblemsolving.com/e/f/f/eff10a32cffaec742403ebfc46ff09bc8ef7b7c5.png)
for all nonnegative integers 
and 
.
make the equation ![\[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\]](http://latex.artofproblemsolving.com/e/7/f/e7fb8c20a43535b5aaed5ab254f1ef043263c62b.png)
true?
be the binary representation of 
. Prove that for any positive integer , at least one of the digits 
, 
, 
, 
equals 
.
, an arbitrary point 
is chosen on the side 
. In triangles 
and 
, the angle bisectors 
and 
are drawn, respectively. The point 
is the circumcenter of 
. Prove that the second intersection point of the circumcircles of triangles 
and 
lies on the line 
.
such that for all 
and for all real numbers 
satisfying 
we have![\[\frac{1}{x_1-x_0} + \frac{1}{x_2-x_1} + \dots + \frac{1}{x_n-x_{n-1}} \geq a \left( \frac{2}{x_1} + \frac{3}{x_2} + \dots + \frac{n+1}{x_n} \right)\]](http://latex.artofproblemsolving.com/8/5/7/857a70e1366e4a5f6eb57056e1c43b7e7eab5ca9.png)
, 
, \dots, 
. So 
, meaning the inequality reads ![\[ \frac{1}{y_1} + \dots + \frac{1}{y_n} \ge a \left( \frac{2}{y_1} + \frac{3}{y_1+y_2} + \dots + \frac{n+1}{y_1+\dots+y_n} \right). \]](http://latex.artofproblemsolving.com/6/6/1/661db672268ceecf63f22531aad0122a1330b275.png)
. First we show this is sharp as 
by taking 
for each 
, thus 
. Then the inequality telescopes as ![\[ 1 - \frac{1}{n+1} \ge a \left( \frac{3}{2} \left( \frac11+\frac12 -\frac{1}{n+1}-\frac{1}{n+2} \right) \right) \]](http://latex.artofproblemsolving.com/e/1/f/e1ffb5d6c160f219a5e025c372d010b73ae932e6.png)
which when forces 
.
which when summed over telescopes to the right thing.. If one takes 
, then we get 
, and when 
one finds the optimal solution is 
for 
. It is hard to solve 
, but one can get things less than 
without too much difficulty. So this gives some sense for where the constant is around.
should grow quickly), so best guess is telescoping sum , which nicely cancels stuff on the right hand side as we then get 
, so 
also telescopes.
is known. However, it seems that it's not easy to guess the value of (though one may know that 
and is around 0.48).
.
. Fix for a while. It's reasonable to apply Cauchy's here. To make sure that the equality holds, suppose that when 
, achieves its maximum (note that is fixed now). Then by Cauchy's,![\[\sum_{k=1}^{i} \frac{b_k^2}{y_k}\geq \frac{(\sum_{k=1}^{i}b_k)^2}{\sum_{k=1}^{i}y_k}......(i\textup{-th inequality})\]](http://latex.artofproblemsolving.com/f/f/b/ffbdad357dec2ac1c562536ae6c8e34b812b000f.png)
-th inequality by 
(here 
) and sum them all:![\[\sum_{i=1}^{n}\frac{1}{y_i}\geq \sum_{i=1}^{n}(\frac{1}{b_i^2}-\frac{1}{b_{i+1}^2})\frac{(\sum_{k=1}^{i}b_k)^2}{\sum_{k=1}^{i}y_k}......(\Delta)\]](http://latex.artofproblemsolving.com/6/7/b/67bb71e5abcc7e33a091797fe18b5854a3ed0801.png)
is non-decreasing. Therefore, we hope that 
.
would be annoying, so it's reasonable to find instead a non-decreasing infinite sequence satisfying![\[(\frac{1}{b_i^2}-\frac{1}{b_{i+1}^2})(\sum_{k=1}^{i}b_k)^2 = (i+1)a\quad\forall i\in\mathbb{N}......(\circ)\]](http://latex.artofproblemsolving.com/b/a/7/ba77a5a29872abc444ecc14682915c983296b8fa.png)
is a positive constant. Then in order to get 
, we have to add 
to it. Here we make a relaxation that is not sharp, so we hope that the difference tends to zero as tends to infinity. In other words, should be unbounded.
, let 
. Then 
. It's natural to guess that 
is a linear function w.r.t. (or first guess that is a polynomial and then deduce that it must be a linear function). With the constraint that 
, it's easy to show that 
.
and![\[b_i = b_1\prod_{k=1}^{i-1}\frac{b_{k+1}}{b_k} = b_1\prod_{k=1}^{i-1}\frac{x_k}{x_{k+1}-1} = b_1\prod_{k=1}^{i-1}\frac{k+2}{k} = \frac{i(i+1)}{2}b_1\]](http://latex.artofproblemsolving.com/2/d/a/2da5e278a332bcd22840bfc9a6112d4f4ced5667.png)
is non-decreasing and unbounded, so is the largest number that satisfies the condition.
, but for now @USJL's motivation might be the best we have., it is fairly straightforward to obtain the equality case by derivatives, and we simply match the corresponding weights after that on the respective Cauchy inequalities.
, such that for any integer 
and any positive real numbers 
the following holds:![\[\sum_{k=1}^{n}\frac{k}{\frac{1}{a_1}+\frac{1}{a_2}+\dotsb +\frac{1}{a_k}} < c\sum_{k=1}^{n}a_k.\]](http://latex.artofproblemsolving.com/9/5/0/95074e1f9caf87d9b814ac3d9d5afdb496b51995.png)
be a real number. Find the smallest real number such that the following inequality holds for all positive integers and positive real numbers 
:![\[\frac{1+t}{a_1}+\frac{2+t}{a_1+a_2}+\dotsb+\frac{n+t}{a_1+\dotsb+a_n} \leq c\left(\frac{1}{a_1}+\frac{1}{a_2}+\dotsb+\frac{1}{a_n}\right).\]](http://latex.artofproblemsolving.com/b/6/f/b6f73ba778685655eaa1c8fdc41f2c672ba8e272.png)
be a real number. Find the smallest real number such that the following inequality holds for all positive integers and positive real numbers :
satisfy the following inequality:
(November, 1998)
.
. By the hypothesis of the problem, we have 
![\[ \sum_{i = 1}^n \frac{1}{a_i} > k \cdot \left( \sum_{i = 1}^n \frac{i + 1}{\sum_{j = 1}^{i} a_j } \right) \]](http://latex.artofproblemsolving.com/a/1/0/a109ee97014daa58f986e70a8dec481593697fed.png)
Any constant 
fails.
Consider the sequence 
for 
. Therefore,![\[ \sum_{i = 1}^n \frac{6}{i(i+1)} > a \cdot \left( \sum_{i = 1}^n \frac{18}{i(i+2)} \right) \]](http://latex.artofproblemsolving.com/6/8/7/68704c1ee37eb9db852236994480dade53e994df.png)
![\[ a < \frac{2}{3} \cdot \frac{1 - \frac{1}{n}}{\frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2}} \]](http://latex.artofproblemsolving.com/2/9/c/29cb2816b473a2f6c526af150a86730fd353faf6.png)
for all 
.![\[ a \le \lim_{n \to \infty} \frac{2}{3} \cdot \frac{1 - \frac{1}{n}}{\frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2}}= \frac{4}{9} \]](http://latex.artofproblemsolving.com/4/4/5/445ecf0ae2526abfe66393d4ef33b85596882961.png)
The constant 
works. Notice that by CS Engel,![\[ \frac{\frac{(i-1)^2}{9}}{a_1 + a_2 + \dots + a_{i - 1}} + \frac{1}{a_i} \ge \frac{ \frac{(i + 2)^2}{9}}{a_1 + a_2 + \dots + a_i} \]](http://latex.artofproblemsolving.com/8/e/d/8ede03f4038feda0278627c108cc827683098935.png)
Therefore, summing all such inequalities gives us![\[ \frac{1}{9a_1} + \sum_{i = 2}^n \frac{1}{a_i} \ge \frac{4}{9} \cdot \sum_{i = 2}^n \frac{i + 1}{a_1 + a_2 + \dots + a_i} \]](http://latex.artofproblemsolving.com/e/2/a/e2a9c2778e1dda57f117fdd67e30388fb20147af.png)
Hence, we have![\[ \sum_{i = 1}^n \frac{1}{a_i} \ge \frac{4}{9} \cdot \sum_{i = 1}^n \frac{i + 1}{a_1 + a_2 + \dots + a_i} \]](http://latex.artofproblemsolving.com/6/2/3/623167ba9411ca9f8ccdc30e4a8ded8c9859c4bb.png)
, which results being an ugly irrational. Then I get 
for 
and 
for 
. To easily apply CS Engel, i expect to have 
for 
. To prove fails, consider the equality case of when .
for all 
with 
. The inequality becomes![\[\sum_{i = 1}^n \frac{1}{t_i} \geq a \left(\sum_{i = 1}^n\frac{i + 1}{t_1 + \ldots + t_i}\right).\]](http://latex.artofproblemsolving.com/4/1/8/418815d01429f63f08cca77085d21f3399662f8b.png)
Since the denominators of terms in the RHS are the partial sums of 's we are motivated to find some bound by Titu's. Indeed, after some time, we arrive at![\[\frac{m^2}{t_1 + \ldots + t_m} + \frac{9}{t_{m + 1}} \geq \frac{(m + 3)^2}{t_1 + \ldots + t_{m + 1}}\]](http://latex.artofproblemsolving.com/0/5/a/05ac8fcd1e6a639f89a078474c8c5fa793de88a0.png)
for all positive integers from 
. This is very good: upon summing over all terms, we get that![\[9\left(\sum_{i = 1}^n \frac{1}{t_i}\right) + \sum_{i = 1}^{n-1} \frac{i^2}{t_1 + \ldots + t_i} \geq \sum_{i = 1}^n \frac{(i + 2)^2}{t_1 + \ldots + t_i} \implies 9\left(\sum_{i = 1}^n \frac{1}{t_i}\right) \geq 4\left(\sum_{i = 1}^{n} \frac{i + 1}{t_1 + \ldots + t_i}\right) \]](http://latex.artofproblemsolving.com/f/b/e/fbe8c119ecb572093bd1543e37bb9d5aeacbd574.png)
showing that the constant 
works. However, it is not yet clear that 
is the largest such constant. In order for equality conditions to hold for all inequalities above, we need![\[\frac{m^2}{9} = \frac{(t_1 + \ldots + t_m)^2}{t_{m + 1}^2}\]](http://latex.artofproblemsolving.com/e/9/3/e933f6587aefd8ca58deb2b6aab26292f8f8fdb1.png)
for all 
. Indeed, it is not particularly hard to verify that the sequence 
verifies these equality conditions, motivated by letting 
. Lastly, we plug this sequence back into the original inequality in 
's. By telescoping, it is clear that![\[\sum_{i = 1}^n \frac{1}{t_i} = 2\left(1 - \frac{1}{n + 1}\right)\]](http://latex.artofproblemsolving.com/3/c/d/3cd281e5e2cb0b461f97d2317481acc33c1610a9.png)
and![\[\sum_{i = 1}^n \frac{i + 1}{t_1 + \ldots t_i} = \sum_{i = 1}^n \frac{i + 1}{\binom{i + 2}{3}} = \sum_{i = 1}^n \frac{6}{i(i + 2)} = 3\left(\frac{3}{2} - \frac{1}{n + 1} - \frac{1}{n + 2}\right)\]](http://latex.artofproblemsolving.com/e/3/d/e3db0cefa4f8d630b8035c85b080a803b112baab.png)
and as tends to infinity the ratio of these two does indeed approach 
hence our desired answer is . 
such that, for all integer 
and for all positive real numbers 
we have
then the ineq fails.
where is a positive integer greater than 
. Then its easy to note that we have 
, now we will put 2 great results that are left to the reader:
Now note that when the ineq becomes:
Thus our claim is proven!. is the maximun, now note that by titu's lemma we have:
Thus we are done 
then the ineq fails. where is a positive integer greater than . Then its easy to note that we have , now we will put 2 great results that are left to the reader:Now note that when the ineq becomes:Thus our claim is proven!. is the maximun, now note that by titu's lemma we have:Thus we are done 
, we need to find a suitable sequence 
. First we thought about the A.P sequence, because the LHS will have common denominator, but the RHS is disgusting. It then reminded us about telescoping sums. What can we substitute in order to make 
and 
to be a telescope-able term? Then we thought for a while and tried the substitution @MathLuis had added, and it worked. So the main idea for the disproval of 
is telescoping sums. Now to prove the other part of the inequality, T2 is natural, because of the denominators in the LHS. Now we send 
to that side and 
remains on the RHS. Now, we want to apply T2 in such a way that we get 
in the numerator of each term in the RHS. So, we want 
and 
and a simple check gives 
. From here, everything is pretty straightforward 
. The construction is 
. We can check that it works. It is inspired by binomial coefficients for ease of computation.
coefficient, it suffices to show 
for positive integers 
. Note that 
for all . We claim that![\[\frac{1}{d_1} + \frac{1}{d_2} + \dots + \frac{1}{d_n} \geq \frac49 \left( \frac{2}{x_1} + \frac{3}{x_2} + \dots + \frac{n+1}{x_n} \right)\]](http://latex.artofproblemsolving.com/1/b/7/1b77ac3335ccb0511b829995925f1f027e54ca99.png)
Observe that by Titu's Lemma, for 
, 
so 
. We have
As desired.
. Let 
, then 
. Then, we have![\[2\sum_{i=1}^{\infty}{\frac{1}{i(i+1)}} \geq 3a \left( \sum_{i=1}^{\infty}{\frac{2}{i(i+2)}} \right)\]](http://latex.artofproblemsolving.com/d/3/9/d39c09768df5ed06936b8961b6ae1c7633370893.png)
so 
.
when 
for 