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.
be real numbers such that ![\[ \frac{1}{x^2-1} + \frac{1}{y^2-1} + \frac{1}{z^2-1} = 1. \]](http://latex.artofproblemsolving.com/8/f/5/8f5d49c9b44e3db3cd85fea7b6521e1dc02d4269.png)
Prove that ![\[ \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} \le 1. \]](http://latex.artofproblemsolving.com/c/0/4/c0438d6eba686d0a11fd4decb4afa9a14aa50bac.png)
such that, for every positive integer 
the integer
is a multiple of 
(Note that 
denotes the greatest integer less than or equal to 
For example, 
and 
)
.
, there exist indices 
such that 
is divisible by .
, find the minimum possible value of 
.
. The point 
is in the interior of . The following ratio equalities hold:![\[\angle PAD:\angle PBA:\angle DPA=1:2:3=\angle CBP:\angle BAP:\angle BPC\]](http://latex.artofproblemsolving.com/9/3/8/9388a91325842f1ec67a4693f0f10af46ed66c6d.png)
Prove that the following three lines meet in a point: the internal bisectors of angles 
and 
and the perpendicular bisector of segment 
.
for which there exists an integer 
such that 
is a divisor of 
.
with 
. The circumcircle 
of has radius 
. There is a point in the interior of the line segment such that 
and the length of 
is . The perpendicular bisector of 
intersects at the points 
and 
. is the incentre of triangle 
.
, where and are relatively prime positive integers. Find 
.
. Hence, we focus on the right side.![\[ U = \left\{ (a,b,c) \mid a,b,c > 0 \text{ and } a^2+b^2+c^2 < 1000 \right\}. \]](http://latex.artofproblemsolving.com/1/d/4/1d4f5748608b93f806d00a36ade8b939c3af8366.png)
This is an intersection of open sets, so it is open. Its closure is ![\[ \overline U = \left\{ (a,b,c) \mid a,b,c \ge 0 \text{ and } a^2+b^2+c^2 \le 1000 \right\}. \]](http://latex.artofproblemsolving.com/2/0/8/208109ca33a19160d3a04e9d8e00b5e97bf7ef40.png)
Hence the constraint set ![\[ \overline S = \left\{ \mathbf x \in \overline U : g(\mathbf x) = 4 \right\} \]](http://latex.artofproblemsolving.com/1/6/0/1600e8cd8aa5219c1b520dc15c03d94547005f3c.png)
is compact, where 
. Excellent.![\[ f(a,b,c) = a^2+b^2+c^2+ab+bc+ca. \]](http://latex.artofproblemsolving.com/b/b/0/bb04a96cd90630a62afe19639303a5842a150e20.png)
It's equivalent to show that 
subject to 
. Over 
, it must achieve a global maximum. Now we consider two cases.
lies on the boundary, that means one of the components is zero (since 
is clearly impossible). WLOG 
, then we wish to show 
for 
, which is trivial.
, we may use the method of Lagrange Multipliers. Consider a local maximum 
. Compute ![\[ \nabla f = \left<2a+b+c, 2b+c+a, 2c+a+b \right> \]](http://latex.artofproblemsolving.com/2/f/1/2f1641cf05573314b51e84dd484ce7ea77820a9c.png)
and ![\[ \nabla g = \left<2a+bc, 2b+ca, 2c+ab\right>. \]](http://latex.artofproblemsolving.com/8/8/9/8894849f0b47c6f236d637bcf3b60d98577a893f.png)
Of course, 
everywhere, so introducing our multiplier yields ![\[ \left<2a+b+c,a+2b+c,a+b+2c\right> = \lambda \left<2a+bc,2b+ca,2c+ab\right>. \]](http://latex.artofproblemsolving.com/3/2/0/32066df167b8e8026f5b37a7142b9bd5fdd9fbd1.png)
Note that 
since 
. Subtracting 
from 
implies that ![\[ (a-b)(\left[ 2\lambda - 1 \right] - \lambda c) = 0. \]](http://latex.artofproblemsolving.com/3/5/c/35c5d53b45364c038d6381f1e58094e85436c39a.png)
We can derive similar equations for the others. Hence, we have three cases.
, then 
, and this satisfies 
.
, 
, 
are pairwise distinct, then we derive 
, contradiction.
. That means 
(of course our conditions force 
). Now ![\[ 2a+2c = a+b+2c = \lambda (2c+ab)
= \frac{1}{2-a} (2c+a^2) \]](http://latex.artofproblemsolving.com/2/f/5/2f5bd28dd016a11506f79087ae947b82c063cb9a.png)
which implies ![\[ 4a+4c-2a^2-2ac = 2c+a^2 \]](http://latex.artofproblemsolving.com/2/5/d/25de76451946469fa8b64a2761d4ff2b737d0a3d.png)
meaning (with the additional note that 
) we have ![\[ c = \frac{3a^2-4a}{2-2a}. \]](http://latex.artofproblemsolving.com/a/7/c/a7c3aa790b033d583e45895f3c7873e5154bb18c.png)
Note that at this point, 
forces 
.
now gives ![\[ \left( 3a^2-4a \right)^2 + a^2 \left( 3a^2-4a \right)\left( 2-2a \right) + \left( 2a^2-4 \right)\left( 2-2a \right)^2 = 0. \]](http://latex.artofproblemsolving.com/4/4/a/44a7145a619354c46158819a02556091be59f9b2.png)
Before expanding this, it is prudent to see if it has any rational roots. A quick inspection finds that 
is such a root (precisely, 
). Now, we can expand and try to factor:
The only root in the interval 
is 
. You can guess what comes next -- write![\[ c = \frac{a(3a-4)}{2-2a} = \frac{1}{8} \left( 7 - \sqrt{17} \right) \]](http://latex.artofproblemsolving.com/5/c/e/5cedfe5a14c0dc14ac2ad8d94c925db853d6f89a.png)
and ![\[ f(a,b,c)
= 3a^2+2ac+c^2
= \frac{1}{32} \left( 121 + 17\sqrt{17} \right)
. \]](http://latex.artofproblemsolving.com/c/9/0/c90708fee647b986cc0fa70de085fb892b6c71a8.png)
This is the last critical point, so we're done once we check this is less than 
. This follows from the inequality 
; in fact, we actually have ![\[ \frac{1}{32} \left( 121 + 17\sqrt{17} \right) \approx 5.97165. \]](http://latex.artofproblemsolving.com/3/4/1/341d40f0bc1f71d45c9667e28a94c264e9b764c1.png)
This completes the solution.
contradiction.
and satisfy ![\[ a^2+b^2+c^2 +abc = 4 . \]](http://latex.artofproblemsolving.com/d/3/a/d3ac01295b077425b5eaae556faa4cd678bce32a.png)
Show that ![\[ 0 \le ab + bc + ca - abc \leq 2. \]](http://latex.artofproblemsolving.com/2/0/9/209b3fcb56c01c335bbcb8d5fb781e28e49a0798.png)
and satisfy ![\[ a^2+b^2+c^2 = 3. \]](http://latex.artofproblemsolving.com/0/b/d/0bd4f8fac6a40a1c5e80d9ec6248b2000a9c185e.png)
Show that ![\[ ab + bc + ca - abc \leq 2. \]](http://latex.artofproblemsolving.com/4/a/3/4a3a07f268e8b8466e2523d60f68e3b69add1452.png)
and 
. Now, obviously 
and:
for , so:
. And we will use this in second equation:
for a triangle . Observe that![\[ab+bc+ca\ge 3\sqrt[3]{a^2b^2c^2}\ge abc\]](http://latex.artofproblemsolving.com/2/3/9/239e232909ba58470014a9c1e61e0ed72d30b59b.png)
because 
, which shows the first part of the inequality. For the latter part, rewrite it as showing![\[3\ge \sum_{\mbox{cyc}}(\cos A+\cos B)^2.\]](http://latex.artofproblemsolving.com/4/1/e/41e879befc0392770f447ed1a603f6a528284066.png)
Note since triangle is not obtuse, we can let 
for some nonnegative 
, so the problem amounts to showing![\[3\ge \sum_{\mbox{cyc}}\left(\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}\right)^2 = 2\sum_{\mbox{cyc}}\frac{x^2}{(x+y)(x+z)}+2\sum_{\mbox{cyc}}\frac{yz}{(y+z)\sqrt{(x+z)(x+y)}}.\]](http://latex.artofproblemsolving.com/1/2/d/12ddbe56c59931fbb6116cbd8b87c807c733a519.png)
By rearranging, the problem amounts to checking![\[6xyz+3\sum_{\mbox{sym}}x^2y\ge 2\sum_{\mbox{cyc}}x^2(y+z)+2\sum_{\mbox{cyc}}yz\sqrt{(x+z)(x+y)}.\]](http://latex.artofproblemsolving.com/e/f/1/ef123180374f03331d46f539b07d85e31d2ef5dc.png)
By cancelling, it is equivalent to show![\[6xyz+\sum_{\mbox{sym}}x^2y\ge 2\sum_{\mbox{cyc}}yz\sqrt{(x+z)(x+y)}.\]](http://latex.artofproblemsolving.com/8/f/5/8f588f9cba0cfae70f6bd57a3c0527beede5f1e5.png)
But by AM-GM, the latter expression is at most![\[\sum_{\mbox{cyc}}(2xyz+y^2z+z^2y),\]](http://latex.artofproblemsolving.com/e/d/c/edca09990ebd2bf8b09c0ab85ad5917b4f4cef26.png)
so done.
as![\[0\le -a((b-1)(c-1)-1)+bc.\]](http://latex.artofproblemsolving.com/a/2/0/a20c7a6d8f25b0ebec703c1beb543cc523330afa.png)
This is clearly true since 
must be nonpositive.
and 
have the same sign. Then 
. Multiplying by and subtracting 
, we get 
. Adding 
![\[c+ab \ge ac+bc+ab-abc.\]](http://latex.artofproblemsolving.com/6/d/c/6dc05a262e3fb7c92e8b3c0c44863a4726d44adb.png)
This means to prove the upper bound it suffices to prove 
. Assume FTSOC that 
. Then,
contradiction. Hence, and we are done.
so wlog any two of 
can be 
or 
and 
so we have ![\[ab+bc+ca-abc=a(b+c)+bc(1-a)\geq 0\]](http://latex.artofproblemsolving.com/c/a/8/ca8cc0695484c910f6dac3780919825cb603e0c9.png)
So lower bound is proved.So now we need to prove upper bound so ![\[a^2+b^2+c^2+abc=4\]](http://latex.artofproblemsolving.com/0/5/0/050bd59f3f1c19aa5159b3b1901a4bb2a47dcdc4.png)
and we know that 
by AM-GM so now put this up![\[a^2+2bc\leq 4\implies bc(2+a)\leq 4-a^2=(2+a)(2-a)\\\implies bc\leq 2-a\]](http://latex.artofproblemsolving.com/5/6/f/56fa77a104934e960055cc4a0fdfd8beaf2fa609.png)
now lets see what we have until now,
for both the cases 
![\[ab+bc+ca-abc\leq ab+2-a+ca-abc=2-a(1-b-c+bc)=2-a(1-b)(1-c)\leq 2\]](http://latex.artofproblemsolving.com/a/1/a/a1a63f4a952705cc03b54b7720def2d27e7d5751.png)
So we are done now !
Equality when 
Q.E.D.
then 
We have to prove 
Equality when 
Q.E.D.
for 
acute. Then the stronger lower bound 
follows from Erdos-Mordell from the circumcenter of to its tangential triangle. has side lengths we can write in terms of by Law of Cosines.![\[abc+2-ab-bc-ca=\sum_{\text{cyc}}bc\cdot\frac{(y-z)^2}{2yz}\]](http://latex.artofproblemsolving.com/e/f/e/efe79336f7e51119827c4f0113e52c8e16c466fe.png)
which is nonnegative so we are done.
is at most 
. For the right side, let denote the open set of all points 
with positive coordinates such that 
, such that its closure 
is given by the set of all points with nonnegative coordinates such that 
.
subject to the constraint 
. By LM, 
achieves a global maximum at some critical point in the constraint set 
.
. Then 
(either directly or if 
), and thus it suffices to show that 
given the condition 
, which is clear.![\begin{align*}
\nabla f &= [2a+b+c, 2b+c+a, 2c+a+b] \\
\nabla g &= [2a+bc, 2b+ca, 2c+ab]
\end{align*}](http://latex.artofproblemsolving.com/0/5/2/052124a799b7ae72539848a4b0b1f8c4b21854bf.png)
and observe that the equation 
implies that 
or 
and cyclic permutations. In particular, either or 
. In this case, the constraint implies 
so 
. along . It remains to verify that ![\[ab+bc+ca-abc = a^2+2a\left(2-a^2\right)-a^2\left(2-a^2\right) - 2 = \left(a-\sqrt 2\right)(a-1)^2 \leq 0\]](http://latex.artofproblemsolving.com/f/e/c/fec792717357d82f90bd5f6001ad2a16c6ef9985.png)
which is clearly true as 
. This completes the proof.
works,
, we want to complete the square, and luckily enough the same trig identity we abused earlier can do just that!
, we multiply both the sides by 
, and we just need to prove -
and 
Prove that![$$3\left(\sqrt[3] 2+\frac{1}{\sqrt[3] 2} -1\right) \geq a+b+c\geq \frac{3+\sqrt{11}}{2}$$](http://latex.artofproblemsolving.com/f/6/6/f66042b4d07212741245953a8940280c94a52988.png)
![$$\frac{3}{2}\left(4+\sqrt[3] 4-\sqrt[3] 2\right) \geq a+b+c+ab+bc+ca\geq \frac{3(1+\sqrt{11})}{2}$$](http://latex.artofproblemsolving.com/0/5/6/056db8bf4d80242d705c265b6a64e063c8fca15a.png)
Prove that
such that 
,
Hence
or
Q.E.D.
and satisfy![\[ a^2 + b^2 + c^2 + abc = 4 .
\]](http://latex.artofproblemsolving.com/0/7/b/07b9ff1d86f856cf428bf6dcff37ae4507907777.png)
Show that![\[ ab + bc + ca - abc \leq 2.
\]](http://latex.artofproblemsolving.com/e/4/1/e414e5bf9bc548b454f6f1e9d77bb4091810c2a6.png)
we have:
