AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
+1 w
be positive real numbers. Prove that:![$$\sqrt[3]{a^3+b^3}+\sqrt[3]{b^3+c^3}+\sqrt[3]{c^3+a^3}\geq \sqrt[3]{2}(a+b+c)$$](http://latex.artofproblemsolving.com/2/6/e/26ebcdae35153bee73abbcfaf4a767d3a32b8912.png)
, 
is the incenter. Line 
intersects circumcircle of at 
, and it is given that 
. 
are points chosen on 
such that 
. Prove that 
be a positive integer. In an 
table, an upright path is a sequence of adjacent cells starting from the southwest corner to the northeast corner such that the next cell is either on the top or on the right of the previous cell. Find the smallest number of grids one needs to color in an table such that there exists only one possible upright path not containing any colored cells.
of positive integers satisfying the following two conditions:
, if 
is in 
, then so are 
and 
; and contains an integer 
such that 
is not divisible by 
.
such that 
and 
are all perfect square.
, let 
be all the positive integers smaller than that are coprime to . Find all 
such that
for all 
.
be an acute triangle. Points 
, and 
lie on a line in this order and satisfy 
. Let 
and be the midpoints of 
and 
, respectively. Suppose triangle 
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
, respectively, such that 
and are concyclic and pairwise different, and 
and are concyclic and pairwise different. Prove that 
and are concyclic.
such that 
and 
.
has perpendicular diagonals. The perpendicular bisectors of the sides and 
meet at a unique point inside . Prove that the quadrilateral is cyclic if and only if triangles 
and 
have equal areas.
such that 
is an integer for every positive integer 
be an integer. Find all positive real solutions to the following system of 
equations:
. The only solution is 
and 
for 
, which clearly works. The key is to sum the equations 
(all sums from here will be from 
to ) ![\[\sum a_{2k-1}=\sum \frac{1}{a_{2k-2}}+\frac{1}{a_{2k}}=2\sum \frac{1}{a_{2k}}=2\sum \frac{1}{a_{2k-1}+a_{2k+1}}\]](http://latex.artofproblemsolving.com/9/1/f/91f01559f102b16f0ac5dad52ca038ce891d3b84.png)
Now using the inequality 
, ![\[\sum a_{2k-1}=2\sum \frac{1}{a_{2k-1}+a_{2k+1}}\leq \frac{1}{2}\sum \frac{1}{a_{2k-1}}+\frac{1}{a_{2k+1}}=\sum \frac{1}{a_{2k-1}}=\sum \frac{1}{\frac{1}{a_{2k-2}}+\frac{1}{a_{2k}}}\]](http://latex.artofproblemsolving.com/d/b/a/dba06922516deb55ab2559c4f88b49833104133c.png)
Now using AM-HM, ![\[\leq \sum \frac{a_{2k-2}+a_{2k}}{4}=\frac{1}{2}\sum a_{2k}=\sum a_{2k-1}\]](http://latex.artofproblemsolving.com/7/8/8/7881f10d9c3fd6ac50fd740ae4ab3922039b318c.png)
Since we have bounded above and below with 
, the inqualities must be equalities!
and 
, which gives the solution.
where 
if 
is odd, 
if is even. Clearly this works.
for all relevant .![$a_{2k} = a_{2k-1} + a_{2k+1} \ge 2\sqrt{a_{2k-1}a_{2k+1}} \ge 2\sqrt{\frac{2}{\sqrt{a_{2k-2}a_{2k}}} \cdot\frac{2}{\sqrt{a_{2k}a_{2k+2}}}} = \frac{4}{\sqrt{a_{2k}}\sqrt[4]{a_{2k-2}a_{2k+2}}} \Rightarrow a_{2k}^3 \sqrt{a_{2k-2}a_{2k+2}} \ge 16$](http://latex.artofproblemsolving.com/9/8/4/984dd6ed25b1e96105b70054e51f5f9ad5f037d8.png)
. Taking product over all relevant , we get 
, thus 
.![$\sum_{k=1}^n a_{2k} \ge n \sqrt[n]{\prod_{k=1}^n a_{2k}} = 2n$](http://latex.artofproblemsolving.com/3/4/9/349ffa209f2d3c70c19dda75b125d86986b02dcf.png)
. So average value of 
is 
.
. Since 
, WLOG let 
. Since 
and , then 
. Since 
and , then 
. Since 
and , then 
. Since 
and then 
. And so on.
, and we have 
. Notice if is odd, then at some point we get 
, 
, and so on, so we get 
, which corresponds to the solution we mentioned at the beginning. is even, then we have 
. And we have 
. But notice that 
. So 
. And so 
.
. Let focus on this last term. We keep the sum 
constant, while reducing 
via smoothing. Notice that if 
and 
is constant, then 
, true by AM-GM, equality holds iff 
. Thus the whenever we have 
we can reduce the value of . The minimum of this expression is then achieved when . And the minimum value of 
is thus 
. But we need this to be the case, for the system of equations to be satisfied. So in fact system of equation can only be solved if . And thus if is even, the only solution is the one mentioned at the beginning as well.
, then either 
or 
, and if WLOG 
then 
and so on.
its identical with the signs switched), so I think it's nontrivial, but couldn't generalize for even.
. Will I get a sympathy point?
, for integer. This works because![\[1 = a_{2k+1} = \frac{1}{a_{2k}} + \frac{1}{a_{2k+2}} = \frac{1}{2} + \frac{1}{2} = 1\]](http://latex.artofproblemsolving.com/e/e/8/ee83b90364ec82e441e2ffc42f33d9ebd571bf15.png)
![\[2 = a_{2k} = a_{2k-1} + a_{2k+1} = 1 + 1 = 2\]](http://latex.artofproblemsolving.com/7/3/d/73d187515563ddd69724dae2db7fcb5d0145bdae.png)
Now we will show that these are the only solutions. Take the indices modulo (so 
). I claim 
. FTSOC, assume 
, and WLOG let 
.![\[(-1)^{k}\left[\frac{1}{a_{2k+4}} - \frac{1}{a_{2n-2k}}\right] = (k+1)(a_{3} - a_{1})+ k(a_{2n} - a_{4}) + (k-1)(a_{6} - a_{2n-2})\]](http://latex.artofproblemsolving.com/b/1/c/b1c2e2431393de793c785491485737ea1cc748e8.png)
![\[+ (k-2)(a_{2n-4} - a_{8}) + (k-3)(a_{10} - a_{2n-6}) + \ldots + (-1)^{k-1}(a_{2n-2(k-1)} - a_{2k+2}) > 0\]](http://latex.artofproblemsolving.com/7/3/2/7321a30f71715fb3f1eff105e06b6b50d1a5065b.png)
, we have![\[\frac{1}{a_{4}} - \frac{1}{a_{2n}} = (a_{3} - a_{1}) > 0\]](http://latex.artofproblemsolving.com/3/3/b/33bbf92700ba0b790c33f9ba2869bde6c1b05fac.png)
This is true by subtracting these two equations:![\[a_{1} = \frac{1}{a_{2}} + \frac{1}{a_{2n}}, a_{3} = \frac{1}{a_{2}} + \frac{1}{a_{4}}\]](http://latex.artofproblemsolving.com/1/9/d/19dd8148a74f30388c941db8b2bdb5b9637db740.png)
, the above formula is true for . Consider the following:![\[a_{2k+2} = a_{2n-2k + 2} + (a_{2k+2} - a_{2n-2k+2})\]](http://latex.artofproblemsolving.com/3/a/f/3af1f4fed024ac701fc7328e74baaf628873dc57.png)
![\[a_{2k+1} + a_{2k+3} = a_{2n-2k+1} + a_{2n-2k+3} + (a_{2k+2} - a_{2n-2k+2})\]](http://latex.artofproblemsolving.com/2/8/d/28d8b8d5dd039dc555660d791c54f27f7c73f720.png)
![\[\frac{1}{a_{2k}} + \frac{2}{a_{2k+2}} + \frac{1}{a_{2k+4}} = \frac{1}{a_{2n-2k}} + \frac{2}{a_{2n-2k+2}} + \frac{1}{a_{2n-2k+4}} + (a_{2k+2} - a_{2n-2k+2})\]](http://latex.artofproblemsolving.com/9/e/4/9e4cc18200b05925799aa12ac6af4cd7dc2ea6d1.png)
If was even, then![\[\frac{1}{a_{2k+4}} - \frac{1}{a_{2n}-a_{2k}} = 2\left[\frac{1}{a_{2n-2k+2}} - \frac{1}{a_{2k+2}}\right] + \frac{1}{a_{2n-2k+4}} - \frac{1}{a_{2k}} + (a_{2k+2} - a_{2n-2k+2})\]](http://latex.artofproblemsolving.com/9/a/1/9a111cb2f5c2a112e25a81d4ffdd96f532b5ab56.png)
By our inductive hypothesis, we get that this is equal to![\[2\left[k(a_{3} - a_{1}) + (k-1)(a_{2n} - a_{4}) + \ldots + (a_{2n-2k+4} - a_{2k})\right]\]](http://latex.artofproblemsolving.com/7/a/e/7ae169fc764f585c099d0cb11933316db45fa8f4.png)
![\[- \left[(k-1)(a_{3} - a_{1}) + (k-2)(a_{2n} - a_{4}) + \ldots + (a_{2k-2} - a_{2n-2k+6})\right] + (a_{2k+2} - a_{2n-2k+2})\]](http://latex.artofproblemsolving.com/8/2/5/825fd1c838eeb78b100445fa5a3cbeb687b1c6fd.png)
![\[= (k+1)(a_{3} - a_{1}) + k(a_{2n} - a_{4}) + \ldots + 2(a_{2n-2k+4} - a_{2k}) + (a_{2k+2} - a_{2n-2k+2})\]](http://latex.artofproblemsolving.com/0/d/2/0d2f5f0f3cb2b35d7479d7804549f0c81a0d437c.png)
This completes the induction for even. We get a similar result for odd. This completes our induction. However, now we have to show that this result is greater than 
. By our inductive hypothesis, if is odd, then 
, so 
. Similarly, for even, 
. Since in the expression, 
, and for each 
for even is greater than , and for each 
for odd is greater than , we conclude this entire expression must be positive. This completes our induction.
, this means 
. If we find some 
such that 
, this will yield a contradiction, since by taking modulo , this means 
are the same variable, a contradiction. However, after simplifying, we get 
, or 
. Taking 
gives the desired contradiction, so it is impossible to have 
.
. By a similar argument, we can show that 
. Let 
. Then, 
, for all . Now,![\[T = a_{1} = \frac{1}{a_{2}} + \frac{1}{a_{2n}} = \frac{1}{2T} + \frac{1}{2T} = \frac{1}{T}\]](http://latex.artofproblemsolving.com/b/7/4/b74e74206889e87e3afbe587530390bc5c087be1.png)
Since 
, this means 
, so are our only solutions.
such that 
has maximal distance from 
If 
then one of the adjacent even values must be less than 
let it be 
Then, we can easily show that 
contradiction. If 
, then one of the adjacent even values must be greater than and we get a similar contradiction. Thus and thus all odd indices are 
so all evens are also and we're done.
and 
and 
and 
, then 
, for 
so 
Also, so 
Then, 
and 
and 
and 
then 
for 
so 
so 
So 
So 
and 
are equal. Then there exist two neighboring terms that are not equal.
so 
gives 
gives 
This is a contradiction, so our assumption was false.
Plugging this into the first equations gives 
and 
Solving gives 
and 
Therefore, our solution is 
![\[a_1^2=\frac{a_1}{a_{2n-1}+a_1}+\frac{a_1}{a_1+a_3}\]](http://latex.artofproblemsolving.com/f/1/4/f14ee379201c65d4135725785458a0934112201d.png)
![\[a_3^2=\frac{a_1}{a_{2n-1}+a_1}+\frac{a_1}{a_1+a_3}\]](http://latex.artofproblemsolving.com/2/9/a/29a3c9667021909506fd089e226110b9b49ba6c4.png)
![\[\cdots\]](http://latex.artofproblemsolving.com/c/e/2/ce26e72ef1985f46ac9b2fedbe0879501540c94c.png)
![\[a_{2n-1}^2=\frac{a_{2n-1}}{a_{2n-3}+a_{2n-1}}+\frac{a_{2n-1}}{a_{2n-1}+a_1}\]](http://latex.artofproblemsolving.com/1/b/b/1bb9a48077063b07e33466a7b107343142cecde3.png)
and summing these gives![\[a_1^2+a_3^2+\cdots+a_{2n-1}^2=n.\]](http://latex.artofproblemsolving.com/a/1/c/a1c7d35813c89dff39057fd07ff5d64809885314.png)
We also have![\[(a_1+a_3+\cdots+a_{2n-1})\cdot(a_1+a_3+\cdots+a_{2n-1}) \]](http://latex.artofproblemsolving.com/8/4/3/843b8d0d0c2680f79069c0db3be36c8b79de7a68.png)
![\[=\left(\frac{2}{a_2}+\frac{2}{a_4}\cdots+\frac{2}{a_{2n}}\right)\left(\frac{a_2}{2}+\frac{a_4}{2}\cdots+\frac{a_{2n}}{2}\right)\]](http://latex.artofproblemsolving.com/4/b/f/4bf8290d11782b69788c876ab10e2f8456b8fbf8.png)
![\[\geq n^2=(1+1+\cdots+1)(a_1^2+a_3^2+\cdots+a_{2n-1}^2).\]](http://latex.artofproblemsolving.com/7/6/f/76f4ab4cfe1f3371922740e8673c879f0025210b.png)
but this is the opposite of Cauchy Schwartz, so all the variables must be equal. It's easy to find the exact solutions from here.
which satisfy 
Observe by cauchy
so the 
add to at least . we have
Proof: For convenience denote 
so by manipulating our relation we can get 
Considering 
as a constant this is a hyperbola in the 
plane with vertices at 
. Since we only care about points in the first quadrant, we only look at the branch with the latter vertex. The claim is now clear. Also note that this means we must have 
, else both vertices of the hyperbola are in the third quadrant and there are no positive solutions. are , so their maxima is 
; WLOG assume 
and for convenience denote 
. So 
and
where we note 
is a decreasing function and the last inequality utilizes the fact that 
.
Expanding the first two equations, we get 
Subtracting 
from the first equation and 
from the second, we obtain 
Similarly, we can show that 
For all 
let 
Then, our equations become 
Now, note that
Since 
we know 
Therefore, 
By multiplying all of our equations, we can show that 
Therefore, 
so by symmetry, 
This implies that 
If is odd, then all 
are equal and all 
are equal, so we must have 
for all 
If is even, then 
so 
and similarly for other for all then 
from which it is easy to see that the only solution is 
for all then 
which yields no solutions.
and 
such that
![\[a_{2k} = \frac{1}{a_{2k - 2}} + \frac{2}{a_{2k}} + \frac{1}{a_{2k + 2}}\]](http://latex.artofproblemsolving.com/6/8/c/68cd6f8f8a912f5f7f5afd75cccb1f66d011f30e.png)
for all 
.
,
as well.![\[a_{2k} \geq \frac{a_{2k - 2} + a_{2k + 2}}{2}.\]](http://latex.artofproblemsolving.com/4/1/0/410fb6718f859e03dadb675a5a6c963fb126bddc.png)
Plugging in from the equation means ![\[\frac{1}{a_{2k - 2}} + \frac{2}{a_{2k}} + \frac{1}{a_{2k + 2}} \geq \frac{\frac{1}{a_{2k - 4}} + \frac{2}{a_{2k - 2}} + \frac{2}{a_{2k}} + \frac{2}{a_{2k + 2}} + \frac{1}{a_{2k + 4}}}{2},\]](http://latex.artofproblemsolving.com/5/9/6/596b57a053b63355128fa72ebf0f643b6e4ee4f6.png)
which simplifies to ![\[\frac{1}{a_{2k}} \geq \frac{\frac{1}{a_{2k - 4}} + \frac{1}{a_{2k + 4}}}{2}.\]](http://latex.artofproblemsolving.com/a/5/2/a5210b73a28c30daf6c0ea708e6c0a385d297cf4.png)
However, since 
for all 
for all 
for 
should not be very large compared to the other terms of this form because the big term gets reciprocated.
for the equations:
.
.
,
.
.
are equal, and this gives 1 2 1 2 ... as the only solution.
ftsoc and wlogging 
work? by inductively spamming the exact same logic over and over you get something like![\[\cdots>a_7>a_3>a_1>a_5>a_9>\cdots\]](http://latex.artofproblemsolving.com/1/b/5/1b5757c34e306a758d38339c40f4d1cb74f0212b.png)
and![\[\cdots>a_{10}>a_6>a_2>a_4>a_8>\cdots\]](http://latex.artofproblemsolving.com/2/8/5/28598b5524eb47d69c3c01e03173eaafe827c8c4.png)
but this seems really sketchy (if its right then by looping around u find everything must be equal which finishes)
to get that all odds are equal to 
?
.
,
and 
is less than 2, since their sum of reciprocals is greater than 1. WLOG that 
.
and 
is less than 1 by a larger margin than 
Therefore, the furthest odd indexed value from 1 is still equal to 1, so all odd values are 1, so all even values are 2. Therefore, the only solution is alternating 1 and 2, which clearly works.
,
since we have 
and 
.
for some 
.
,
and 
must be true. Assume WLOG that 
and 
,
giving 
,
.
to this inequality gives 
.
,
,
,
.
holds. At this point, the argument can be repeated from the beginning with 
replaced by 
;
if 
,
if 
,
if 
,
if 
.
,
,![\[n < \sum_{i=1}^{n/2} a_{2x + 4i} = \sum_{i=1}^{n/2} a_{2x + 4i - 1} + a_{2x + 4i + 1} = \sum_{i=1}^n a_{2i-1} = \sum_{i=1}^{n/2} a_{2x + 4i + 1} + a_{2x + 4i + 3} = \sum_{i=1}^{n/2} a_{2x + 4i + 2} < n,\]](http://latex.artofproblemsolving.com/8/e/f/8ef6c6e2ce59317be1e6822215ff5a34066fdf69.png)
contradiction. 
,
for all 
for all 
![\[ \sum_{k \ \text{odd}} a_k = 2 \left(\sum_{k \ \text{even}} \frac{1}{a_k} \right) = 2 \left(\sum_{k \ \text{odd}} \frac{1}{a_k + a_{k+2}}\right) \le \frac{1}{2} \left(\sum_{k \ \text{odd}} \frac{1}{a_k} + \frac{1}{a_{k+2}}\right) = \sum_{k \ \text{odd}} \frac{1}{a_k} = \sum_{k \ \text{even}} \frac{1}{\tfrac{1}{a_k} + \tfrac{1}{a_{k+2}}} \le \sum_{k \ \text{even}} \frac{a_k+a_{k+2}}{4} = \frac{1}{2} \sum_{k \ \text{even}} a_k. \]](http://latex.artofproblemsolving.com/f/2/8/f2868f1bc97189401e473c4c27fd3fe98ddfd52b.png)
Realize that by summing the second column of equations, ![\[ \sum_{k \ \text{even}} a_k = 2 \sum_{k \ \text{odd}} a_k, \]](http://latex.artofproblemsolving.com/6/5/9/6590c7e87a34b2de59ac901956629e06c8d489fe.png)
and that our previous one-liner implies the ![\[ \sum_{k \ \text{even}} a_k \ge 2 \sum_{k \ \text{odd}} a_k. \]](http://latex.artofproblemsolving.com/0/7/9/079255c87a4271ef013d5d9d1cfbc15801d0e94d.png)
Thus all inequalities in the one-liner are tight, which implies 
for all even 
is a constant sequence (suppose that all terms are 
)
is also a constant sequence (with all terms 
)
so 
.
for odd 
for even 
so 
Thus consider the maximum 
It must be 
as otherwise we have 
and taking indices cyclically, we see one of 
is 
so assume without loss of generality that it is 
Then we have 
if these are all positive. However, when 
this is positive when 
and we have 
contradicting maximality finishing.
![\[a_{2k} = a_{2k-1}+a_{2k-3} = \frac{1}{a_{2k-2}}+\frac{2}{a_{2k}}+\frac{1}{a_{2k+2}}\]](http://latex.artofproblemsolving.com/1/7/8/1788111f9395e836abc2cf643d8d63b108eb5ee3.png)
Let 
be the minimum of all ![\[m - \frac{2}{m} = \frac{1}{a_{2k-2}}+\frac{1}{a_{2k+2}} \ge \frac{2}{M}\]](http://latex.artofproblemsolving.com/f/5/a/f5a880b132bec7fe00c199d0cd2cca30747f914f.png)
![\[M-\frac{2}{M} = \frac{1}{a_{2\ell-2}}+\frac{1}{a_{2\ell+2}} \le \frac{2}{m}\]](http://latex.artofproblemsolving.com/d/3/b/d3b25901bb4d52b77416783d7cd12aaabf9eb67e.png)
This implies 
,
.
and 
.
for all odd 
.
and get 
.
.
,
(with indices taken mod 2n). However, if 
for all odd 
.
and 
.
only. To see this, reciprocating the second equation, we have 
,![\[a_1+a_3+\cdots+a_{2n-1} = 2\left(\frac 1{a_2}+\frac 1{a_4} + \cdots+\frac 1{a_{2n}}\right) = 2\left(\frac 1{a_1+a_3}+\frac 1{a_3+a_5}+\cdots+\frac 1{a_{2n-1}+a_1}\right).\]](http://latex.artofproblemsolving.com/c/c/8/cc8ac12c5ea956bb64f9e5c2847c2690920df822.png)
On the other hand, by reciprocating the first equation, we have ![\[\frac 2{a_{2i-1}} = \frac 2{\frac 1{a_{2i}}+\frac 1{a_{2i-2}}} \leq \frac{a_{2i}+a_{2i-2}}2.\]](http://latex.artofproblemsolving.com/1/4/4/14481f76ca3cc78f60bc85649974eb34a01a8bd9.png)
Summing these equations then yields
By AM-HM we have 
,
and 
and the result follows.
.
are constant as well.
and let 
.
And.
But we have that.
While we also have that.
Hence, we have that 
implying that the sequence 
is constant, which means that 
for all 
for all 
Notice that summing all of the equations on the left gives that
we have 
with equality if and only if 
or 
which is obviously true.
Now,
This seems to be useless. However, it means that all of our inequalities must be equalities. So, 
and 
by the equality case of the claim. We get 
and 
so 
and 
Hence, we get 
and 
,
,
for all 
where indices are taken 
.![\[2b_k = \dfrac{1}{b_k} + \dfrac{1}{2b_{k - 1}} + \dfrac{1}{2b_{k + 1}} \ge \dfrac{2}{\sqrt[4]{b_k^2 b_{k - 1} b_{k + 1}}} \implies b_k^6 b_{k - 1} b_{k + 1} \ge 1\]](http://latex.artofproblemsolving.com/b/5/8/b5835d4b029f96440148a0d54cb6552f90181438.png)
which multiplying cyclically gives 
.![\[b_1 + b_2 + \dots + b_n = \dfrac{1}{b_1} + \dfrac{1}{b_2} + \dots \dfrac{1}{b_n} \le \left( \dfrac{1}{b_1} + \dfrac{1}{b_2} + \dots \dfrac{1}{b_n} \right) (b_1 b_2 \dots b_n)^{2/n} = \sum_{\text{sym}} b_1^{2/n - 1} b_2^{2/n} \cdots b_n^{2/n}.\]](http://latex.artofproblemsolving.com/d/1/c/d1cc8752a6162641904d992a8ed960e238d1e731.png)
But 
so by Muirhead the opposite of the above inequality must hold. Then equality occurs everywhere, thus 
for all 
.
by summing the equations on the right. We also have 
where the inequality holds by AM-HM. This means that we must have the equality case, implying that 
.
T
.![\[\sum a_{2i-1}=\sum\tfrac{2}{a_{2i}}=\sum\tfrac{2}{a_{2i-1}+a_{2i+1}}\le\sum\tfrac{1}{a_{2i-1}}=\sum\tfrac{a_{2i-2}a_{2i}}{a_{2i-2}+a_{2i}}\le\sum\tfrac{a_{2i}}{2}=\sum a_{2i-1},\]](http://latex.artofproblemsolving.com/9/a/d/9ad521c44283b8db42621b96eb20b908611e7717.png)
with equality iff both 
and 
,
.
for all 
for all 
for all 
for all 
)
,
.
the larger root, so 
.
,
,
and 
.
,