AoPS Academy
, and 
be positive real numbers, such that 
. Prove the inequality![\[\frac{x^3}{2 + x} + \frac{y^3}{2 + y} + \frac{z^3}{2 + z} \ge 1.\]](http://latex.artofproblemsolving.com/b/e/5/be5819a67c3cd78f2dea35fdccf48688c720ce3c.png)
When does the equality hold?
such that there exists a pair 
of positive integers, such that 
is not divisible by the cube of any prime, and 
![$f : \mathbb{Q}[\sqrt{2}] \to \mathbb{Q}[\sqrt{2}]$](http://latex.artofproblemsolving.com/c/a/1/ca172b912bf5bf7178cbbd5745fbc0a6a8b27503.png)
such that for all ![$x, y \in \mathbb{Q}[\sqrt{2}]$](http://latex.artofproblemsolving.com/9/f/e/9fede0b2fd22836372b8afc488ccd8a14db5d0c2.png)
,
where ![$\mathbb{Q}[\sqrt{2}] = \{ a + b\sqrt{2} \mid a, b \in \mathbb{Q} \}$](http://latex.artofproblemsolving.com/f/a/c/fac8bf68761534a139d570bf66b1aa5dfb59e8db.png)
.
be a circle, 
and 
are parallel chords and 
is a in its middle point 
and 
. 
is the middle of 
. Prove that 
with orthic triangle 
, let 
, 
, 
be points such that 
. Let 
be the trilinear polar of the polar conjugate of wrt , and define 
and 
cyclically. Let 
be the triangle bounded by , , . and are bilogic.
, a function 
is called interesting if the following two conditions hold:
such that 
., we have 
(where 
denotes the -th iteration of 
).
.
for which there is no interesting function
such that Bob can always find the two coins in at most guesses?
, 
, 
, 
, 
, 
. In general, 
. Thus, we have the inequality
The summation telescopes into 
, so we have 
, as desired.
divides only if is prime. Suppose the smallest divisor of other than 
is 
. Obviously, is prime. Note also that is the smallest divisor of 
other than . Then, we have that 
and 
. We then have that our sum is equal to
However, if 
, then the LHS of the inequality is an integer, contradicting the minimality of . Thus, the sum cannot divide if is composite, as desired. 
for all 
.
has![\[ D = n^2 (\frac{1}{d_{1}d_{2}} + \frac{1}{d_{2}d_{3}} + \cdots +
\frac{1}{d_{k-1}d_{k}}) \]](http://latex.artofproblemsolving.com/1/9/8/1984aaa1fbc73c91590a7f765d4bbf418f1c1c8c.png)
Since 
, we have 
for all .![\[ \frac{1}{d_{1}d_{2}} + \frac{1}{d_{2}d_{3}} + \cdots +
\frac{1}{d_{k-1}d_{k}}
\leq \sum_{i = 1}^{k-1}\frac{1}{(i)(i+1)}
= \sum_{i = 1}^{k-1}\frac{1}{i} - \frac{1}{i+1}\]](http://latex.artofproblemsolving.com/c/3/3/c33e1fd69034ccf3266c447013e526893e123f07.png)
By telescoping, the above sum is equal to 
, so 
.
if and only if is prime. is prime, 
, and trivially 
. is not prime and proceed by contradiction. If , then 
for some positive integer . By the above, cannot be one, so it has to be 
the smallest non-one divisor of . This is the smallest prime in the prime factorization of , which is also the smallest prime in the prime factorization of , and therefore 
.
. Equality is only when 
is the only term in 
, i.e. 
, implying that is prime, which is a contradiction. Otherwise 
, contradicting 
.
Thus, we just need to show that 
Indeed, we know that 
, 
, 
, so it follows that
and the desired inequality follows.
divides if and only if is a prime. are 
, so the given sum is equal to 
. do not work, let be the smallest prime that divides . Then, , and 
. For to divide , we must have 
for some positive integer 
. Since is composite and has at least 
positive divisors, we must have 
which is equivalent to 
. However, since is the smallest prime that divides , is also the smallest prime that divides , so such cannot exist, and we reach a contradiction if is composite. If is prime, then we could have 
as then we only have 
divisors, and the condition is satisfied as shown above. 
, 
, 
so 
For the second part, we claim the answer is when is prime. Let be the smallest prime factor of . If 
, then 
, so prime work. If is not prime, and suppose 
. Since is the smallest prime dividing , we have 
. But 
so 
, contradiction. Therefore, only prime work.
, hence we have that.
. is a prime. is a prime, this case is trivial. is composite, then we have that.
Where is the smallest prime of .
where 
is a divisor of .
, hence only prime works.
, 
, 
, , 
. Observe that:
Now, since 
, we are done. is composite, we have 
, where is the smallest prime divisor of . Observe that 
, however this is the largest divisor of , so we have a contradiction. If is prime, it works.
, as desired. We claim it is a divisor of only when is prime. It is easy to see all prime numbers work. To see that nothing else works, take the smallest prime divisor of as , clearly the second largest divisor of is 
, but 
, so we must have this being the only term in the summation, so forces prime.
Notice that: 
where is a the smallest prime factor of . Then, notice that: 
is the largest divisor of . Thus, it should be the only term in the summation: 
inorder to have 
. Therefore, 
which implies is a prime and we are done.
The second part is only true when 
for some prime . This works because 
. If was compoite, then consider the smallest prime dividing it. The desired sum is then strictly greater than 
, implying the desired contradiction.
is less than where 
varies from to 
. Note that 
is less than or equal to
with equality holding iff 
. Now obviously, for each there must be atleast one pair of consecutive divisors with a difference greater than so the above sum is less than 
which is less than . is 
but this is so the given sum is greater than this unless there is only one term in the sequence, that is, . This implies must be a prime and all primes can easily be verified to work. So the answer to the second part is must be a prime.
is a prime, 
and 
are all the divisors of . Then,![\[d_1d_2=n \nmid n^2\]](http://latex.artofproblemsolving.com/0/b/4/0b490114cf04897f20cc482101d77a603519c478.png)
so the inequality holds with a divisor of . is composite. Then, 
. Now note,![\[d_1d_2+d_2d_3 + \dots + d_{k-1}d_k = \frac{n^2}{d_kd_{k-1}}+\frac{n^2}{d_{k-1}d_{k-2}}+\dots + \frac{n^2}{d_2d_1}\]](http://latex.artofproblemsolving.com/7/b/8/7b8a8604f3ec81d99f6a1fb34e4c15c27e5f24c6.png)
First note,![\[\frac{n^2}{d_kd_{k-1}}+\frac{n^2}{d_{k-1}d_{k-2}}+\dots + \frac{n^2}{d_2d_1} > \frac{n^2}{d_2d_1} = \frac{n^2}{d_2}\]](http://latex.artofproblemsolving.com/f/2/f/f2f5be8a9d28f5daf8889e7f1eefbd1610791ee4.png)
Further, since clearly 
for all 
,![\[\frac{n^2}{d_kd_{k-1}}+\frac{n^2}{d_{k-1}d_{k-2}}+\dots + \frac{n^2}{d_2d_1} \le \frac{n^2}{k(k-1)}+\frac{n^2}{(k-1)(k-2)}+\dots + \frac{n^2}{2 \cdot 1} = n^2 \left(1- \frac{1}{k}\right) < n^2\]](http://latex.artofproblemsolving.com/6/6/0/660f3eb3d9a1ea4c87281cafa62a66edb163c619.png)
Since 
is the smallest divisor of , it must be prime and thus, it is also the smallest divisor of (every prime divisor of is a prime divisor of ). Thus, is the largest divisor of less than . But since,![\[\frac{n^2}{d_2}<\frac{n^2}{d_kd_{k-1}}+\frac{n^2}{d_{k-1}d_{k-2}}+\dots + \frac{n^2}{d_2d_1} < n^2\]](http://latex.artofproblemsolving.com/8/d/a/8da6bfb1748b55e1c34f9d1b7511ffb690f44c35.png)
it cannot be a divisor of .
Now if is composite 
so 
so this cannot possibly divide 
But if is prime it clearly works. Hence that is the answer.
for all 
Hence, 
For equality, observe that 
Hence, 
but it also can't be since we proved earlier it is less than Hence, has 2 prime factors. This obviously works.
Prove that
Where 
,
be a positive integer, with divisors 
.
when 
![\[ S=\sum d_id_{i+1}<n^2\iff \sum\frac{1}{d_id_{i+1}}<1\]](http://latex.artofproblemsolving.com/3/b/2/3b211e0d1666d5dacc2f354c78b781c7e3954002.png)
![\[ \sum\frac{1}{d_id_{i+1}}\leq\sum\frac{d_{i+1}-d_i}{d_{i+1}d_i}=\sum\left(\frac{1}{d_i}-\frac{1}{d_{i+1}}\right)=1-\frac{1}{n}<1\]](http://latex.artofproblemsolving.com/e/8/1/e81e4f453c0ac5df88f69e112877b9a562e1e9b2.png)
.
is a divisor of 
also. Since the smallest non trivial divisor of 
which is 
,
.
and therefore 
,
and 
we will have
.
,
,
,
,
and 
,
and so the above is true.
.
,
)
then there exists a 
.
by the first part and so 
.
,
and we get 
which is false because 
.
.
if and only if 
for every i
so clearly this inequality holds
)
and equality holds only when n is a prime so thats the only posible case
.
.
.
we know 
and 
where 
,![\[ d_id_{i+1} = \frac{n^2}{d_{k+1-i}d_{k-i}} = \frac{n^2}{d_{k+1-i}-d_{k-i}}\left(\frac{1}{d_{k-i}} - \frac{1}{d_{k+1-i}}\right) \le n^2\left(\frac{1}{d_{k-i}} - \frac{1}{d_{k+1-i}}\right), \]](http://latex.artofproblemsolving.com/8/4/8/8487980413d9f0f96c57231700b84b2759de0570.png)
since 
.![\[ \sum_{i=1}^n d_id_{i+1} \le n^2 \sum_{i=1}^n \frac{1}{d_{k-i}} - \frac{1}{d_{k+1-i}} = n^2\left(\frac{1}{d_1}-\frac{1}{d_k}\right)=n^2(1-1/n) < n^2.\]](http://latex.artofproblemsolving.com/3/4/7/347a5f5e7be57ba575889f92ba4c5c9daad0b8e5.png)
(b) Let 
.
.
,
.
is the largest proper divisor of 
,
.
,
,
if and only if 
This proves the first part.
)
)![\[ d_k d_{k-1} + d_{k-1} d_{k-2} + \dots + d_2 d_1 > d_k d_{k-1} = \frac{n^2}{p} \]](http://latex.artofproblemsolving.com/9/f/9/9f94e65ff5fdd58bc6d8981d02ace08eb0a704d9.png)
exceeds the largest proper divisor of 
so the first assertion is true. Observe that 
where 
then 
contradicting that this quotient is a divisor of 
.
![\[S:=d_1d_2+\cdots+d_{k-1}d_k=n^2\left(\frac{1}{d_1d_2}+\cdots+\frac{1}{d_{k-1}d_k}\right)\le n^2\left(\left(\frac{1}{d_1}-\frac{1}{d_2}\right)+\cdots+\left(\frac{1}{d_{k-1}}-\frac{1}{d_k}\right)\right),\]](http://latex.artofproblemsolving.com/0/6/a/06a48f00face21b335a14b5e17e8b9f64993b0d7.png)
so![\[S\le n^2(1-1/n)=n^2-n<n^2,\]](http://latex.artofproblemsolving.com/9/5/6/9568b632cb19c955f87e6bc2041bf1c74bbe9823.png)
as desired.
where 
.
,
.![$[x,y]$](http://latex.artofproblemsolving.com/3/d/3/3d3692c09209718827eeb3f7fcd218f7c92ad98e.png)
,
will be at the edges, so![\[x^2+xy\le xz+zy\le y^2+xy,\]](http://latex.artofproblemsolving.com/e/2/a/e2a6f29f65e8763896ce4b67f47d61aea9924d73.png)
so certainly 
Using the lemma, we see that![\[d_ad_b<d_ad_{a+1}+d_{a+1}d_{a+2}+\cdots+d_{b-1}d_b,\]](http://latex.artofproblemsolving.com/3/d/9/3d990e926776bc5a102ce6471dc18098c11cf502.png)
so 
.
,
is if 
.
.
and so on.
.
must be 
,
but 
if there are more than 
can't divide 
we get it suffices to prove
Now note that
For the other part, note that
But 
.
which divides 
if