AoPS Academy
+2 w
has incentre 
and incircle touching 
at 
. Let 
be the antipode of 
in the circumcircle of . Point 
is chosen on the internal angle bisector of 
such that 
. Let 
be the midpoint of arc 
, and let 
be the midpoint of 
. Prove that 
x square board has the row and column numbered from 
to , square in 
row and 
column get symbolized by square 
. Subset of squares on the board is called "good" subset if two random squares 
belong to satisfy that the squares 
with 
or 
are not belong to . Find the minimum number of "good" distinct subsets such that each square on the board belongs to only one subset
+1 w be a triangle and let and 
denote its incentre and circumcentre respectively. Let 
be the circle through 
and 
which is tangent to the incircle of the triangle ; the circles 
and 
are defined similarly. The circles and meet at a point 
distinct from ; the points 
and 
are defined similarly. Prove that the lines 
and 
are concurrent at a point on the line 
.
+1 w
defined as follows:![\[
a_1 = \frac{\sqrt{6}}{3},
\quad
a_{n+1} = a_n + \frac{1}{3a_n},
\quad
\forall n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/8/9/2/892a2676214a3c837d6b18a10a4f6f83bc099f74.png)
![\[
0 \le a_n \sqrt{6} - 2\sqrt{n}
\le
\frac{1}{4\sqrt{n}}
\Bigl(
1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}
\Bigr),
\quad
\forall n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/f/0/8/f0824edd96ddfeaa107d729466134812ad7ea777.png)
b,For each 
, define![\[
b_n = \frac{3a_n^2 - 2n - 1}{\ln(n+1)}.
\]](http://latex.artofproblemsolving.com/a/9/3/a937b379140df5aa51d7ede1a685d263c3229e96.png)
Compute the limit 
, with help of 
and Moivre's formula.
. Easy to guess that the answer is , but I haven't found the reason yet. :(
houses on a straight street are one or more boys. At what point should all the boys meet so that the sum of the distances that they walk is as small as possible?
![\[
\text{Find: }
10. \sum_{n=1}^\infty \sum_{m=-\infty}^\infty \frac{1}{n^p m^2 (m^2 + 1)^3 (n+1)^q}, \quad 2 \leq p, q \in \mathbb{Z},
\]](http://latex.artofproblemsolving.com/f/3/3/f3360b85ce092804a9de263cac49aed025ea8cee.png)
![\[
11. \sum_{n=1}^\infty \sum_{m=-\infty}^\infty \frac{(-1)^{n+m}}{n^p m^2 (m^2 + 1)^3 (n+1)^q}, \quad 2 \leq p, q \in \mathbb{Z}.
\]](http://latex.artofproblemsolving.com/9/4/5/945aaa2340a4fb314e5eb218ba5286eb054c1b6f.png)
![\[12.
\int_{0}^{\frac{\pi}{4}} \frac{\sin x}{\cos(2x) + 2} \tan^{-1}\left(\frac{\cos x \cot(2x)}{\sqrt{2}}\right) dx
= \frac{5\pi^2}{48\sqrt{2}} - \frac{\pi}{4\sqrt{2}} \cos^{-1}\left(\frac{1}{3}\right).
\]](http://latex.artofproblemsolving.com/b/1/3/b13a5b1c537926c8323170d3d533b0bc71bb06a1.png)
, because I was in Spain yet as a contestant.![\[\sqrt{x^2+y^2} \leq \sqrt{2}(x+y-\sqrt{xy}) \quad \forall x ,y >0 \eqno (*),\]](http://latex.artofproblemsolving.com/e/0/0/e0016a52c1c81ea8454af914135e3eb2eed98061.png)
which is clearly.![\[\sqrt{\frac{x^2+y^2}{x+y}} \leq \frac{\sqrt{2}(x+y-\sqrt{xy})}{\sqrt{x+y}}=\sqrt{2(x+y)}-\frac{\sqrt{2xy}}{\sqrt{x+y}} \quad \forall x ,y >0. \]](http://latex.artofproblemsolving.com/e/6/0/e60c42a9ba24178a29e6647defd9c315887054a4.png)
![\[\sqrt{\frac{a^2+b^2}{a+b}}+\sqrt{\frac{b^2+c^2}{b+c}}+\sqrt{\frac{c^2+a^2}{c+a}} \leq \sqrt{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})-\sqrt{2}\left( \frac{\sqrt{ab}}{\sqrt{a+b}}+ \frac{\sqrt{bc}}{\sqrt{b+c}}+\frac{\sqrt{ca}}{\sqrt{c+a}} \right). \]](http://latex.artofproblemsolving.com/2/4/e/24ebb48b038c45051534cfdb93ac5a9d67712766.png)
![\[\frac{\sqrt{ab}}{\sqrt{a+b}}+ \frac{\sqrt{bc}}{\sqrt{b+c}}+\frac{\sqrt{ca}}{\sqrt{c+a}} \geq \frac{3}{\sqrt{2}}.\]](http://latex.artofproblemsolving.com/f/a/c/fac6547f2461134f07737c20ad9c809d4322dfb4.png)
и 
We have 
![\[
\frac{\sqrt{ab}}{\sqrt{a+b}}+ \frac{\sqrt{bc}}{\sqrt{b+c}}+\frac{\sqrt{ca}}{\sqrt{c+a}}=\frac{1}{\sqrt{\frac{1}{a}+\frac{1}{b}}}+
\frac{1}{\sqrt{\frac{1}{b}+\frac{1}{a}}}+\frac{1}{\sqrt{\frac{1}{c}+\frac{1}{a}}}=\frac{1}{\sqrt{x+y}}+
\frac{1}{\sqrt{y+z}}+\frac{1}{\sqrt{z+x}} \geq \\
\left[ AM-HM \right] \geq \frac{9}{\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}} \geq \left[ AM-QM \right] \geq
\frac{9}{\sqrt{3((x+y)+(y+z)+(z+x))}}=\\
\frac{9}{\sqrt{6(x+y+z)}}=\left[ x+y+z \leq 3 \right] \geq \frac{9}{\sqrt{6 \cdot 3}}=\frac{3}{\sqrt{2}},
\]](http://latex.artofproblemsolving.com/a/f/c/afc21bb50cd2825987d0d547f8067c90a4664904.png)
, which is true by AM-QM.
, 
, and 
. We wish to show that if 
, then
, the 1st power mean of these numbers is greater than or equal to the -2nd power mean, whence 
. Rearrange this to get
, 
. Furthermore, observe that 
. Applying these two inequalities gives
![\[\sqrt{2} \cdot \sqrt{a+b}=\sqrt{2} \cdot \sqrt{\frac{(a+b)^2}{a+b}}=\sqrt{2} \cdot \sqrt{\frac{a^2 + b^2}{a+b}+\frac{2ab}{a+b}} \ge \sqrt{\frac{a^2 + b^2}{a+b}}+\sqrt{\frac{2}{\frac{1}{a}+\frac{1}{b}}} \]](http://latex.artofproblemsolving.com/a/e/6/ae6c797b1c36f06f67a0443ad47234dcfb648efd.png)
![\[\sqrt{\frac{2}{\frac{1}{a}+\frac{1}{b}}} +\sqrt{\frac{2}{\frac{1}{b}+\frac{1}{c}}} +\sqrt{\frac{2}{\frac{1}{c}+\frac{1}{a}}} \ge \frac{9\sqrt{2}}{\sqrt{\frac{1}{a}+\frac{1}{b}}+\sqrt{\frac{1}{b}+\frac{1}{c}}+\sqrt{\frac{1}{c}+\frac{1}{a}}} \]](http://latex.artofproblemsolving.com/1/6/d/16d965335847d9d560438faeb2dec84d7a8987c2.png)
![\[\frac{9\sqrt{2}}{\sqrt{\frac{1}{a}+\frac{1}{b}}+\sqrt{\frac{1}{b}+\frac{1}{c}}+\sqrt{\frac{1}{c}+\frac{1}{a}}} \ge \frac{9\sqrt{2}}{\sqrt{\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}}} \ge 3\]](http://latex.artofproblemsolving.com/5/a/a/5aa487b3205b1b3772d2ea8518ab39fb26022df3.png)
.
and by inequality (2),![\[ \sqrt{\frac{a^{2}+b^{2}}{a+b}}+\sqrt{\frac{b^{2}+c^{2}}{b+c}}+\sqrt{\frac{c^{2}+a^{2}}{c+a}}+3\leq\sqrt{2}\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right) \quad \blacksquare \]](http://latex.artofproblemsolving.com/7/d/d/7dd241bef190e7feb2f60e47181c040ae5fa3789.png)
, 
, 
be positive real numbers such that 
. Prove that![\[\sqrt{\frac{a^2+b^2}{a+b}}+\sqrt{\frac{b^2+c^2}{b+c}}+\sqrt{\frac{c^2+a^2}{c+a}}+3\leq \sqrt{2}\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)\]](http://latex.artofproblemsolving.com/d/f/4/df489a09a27cee10ce445c6454e4c35a1b383a28.png)
we have 
Write 
so our given inequality is equivalent to 
Apply Holder's inequality to get 
However, since 
we get the desired bound. Equality occurs when 
. 
etc., then 
. We wish to show ![\[
\sum \sqrt{2(a+b)} - \sqrt{\frac{a^2+b^2}{a+b}} \geq 3 \iff \sum \frac{\sqrt 2(a+b)-\sqrt{a^2+b^2}}{\sqrt{a+b}} \geq 3 \iff \sum \frac{\sqrt{2} (x+y) - \sqrt{x^2+y^2}}{\sqrt{xy(x+y)}} \geq 3.
\]](http://latex.artofproblemsolving.com/3/6/b/36b3b94a8f229a5cce3905a63f1d580b4b878f03.png)
Note that multiplying 
by 
does not increase the left hand side, so assume 
. Now, we claim that in fact ![\[
\sqrt 2(x+y)-\sqrt{x^2+y^2} \geq \sqrt{2xy}.
\]](http://latex.artofproblemsolving.com/4/8/7/487f6d539970e8c38d4707789aa951dff2f6be7e.png)
Indeed, let 
for 
, then we want ![\[
\sqrt{2}(u^2+v^2-uv) \geq \sqrt{u^4+v^4},
\]](http://latex.artofproblemsolving.com/0/e/d/0edf4ad62b9a011a59a7f204bdfbc672e13e9539.png)
or equivalently ![\[
\sqrt 2((u-v)^2+uv) \geq \sqrt{u^4+v^4} \iff 2((u-v)^2+uv)^2 \geq u^4+v^4 \iff 2(u-v)^2((u-v)^2+2uv) \geq (u^2-v^2)^2 = (u-v)^2(u+v)^2.
\]](http://latex.artofproblemsolving.com/b/9/0/b9080d70f3c3330df3ab3653923ddefafd0c8504.png)
If 
we are clearly done, else we want to show 
which is clear. So, it suffices to show ![\[
\sum \frac{\sqrt 2}{\sqrt{x+y}} \geq 3\iff \sum \sqrt{\frac{2}{3-x}} \geq 3.
\]](http://latex.artofproblemsolving.com/c/0/5/c0567a131847a84820c72c2ed4b623ca48063239.png)
However, 
is convex with second derivative 
, so we have ![\[
\sum \frac{\sqrt 2}{\sqrt{3-x}} \geq \frac{\sqrt 2\cdot 3}{\sqrt{3-1}} = 3
\]](http://latex.artofproblemsolving.com/b/8/4/b84036f96e9da3fd6c1e754cb430f5751bb9c106.png)
as desired. 
.
. Luckily, this is easy by Holder:
, 
, and 
. Then, 
, so the inequality is equivalent to
or 
This rearranges to 
By QM-AM, we get 
, which implies 
. Therefore, we get that the left hand side is equal to
This means that by Holder's Inequality, we have
which means 
, so ![\[\sqrt{\frac{a^2+b^2}{a+b}}\le \sqrt{2}\left(\sqrt{a+b}-\sqrt{\frac{2ab}{a+b}}\right)\]](http://latex.artofproblemsolving.com/3/e/e/3ee346765805d090c99d7a7ff08913e4b7476cd3.png)
so it remains to show that 
![\[\left(\sum_{cyc}\sqrt{\frac{2ab}{a+b}}\right)^2(\sum_{cyc}\frac{a+b}{2ab})\ge 27\]](http://latex.artofproblemsolving.com/8/d/e/8deb0d9bb7a1f95bba3ae14e816f9a51742c0fe7.png)
, but 
which implies the result.
![\[3\sqrt{3} \cdot \sqrt{\frac{abc}{ab + bc + ca}} + \sum_{\mathrm{cyc}} \sqrt{\frac{a^2 + b^2}{a + b}} \le \sqrt{2}(\sqrt{a + b} + \sqrt{b + c} + \sqrt{c + a}).\]](http://latex.artofproblemsolving.com/7/6/2/762bca1a7cf17c37c625774d0596a337e5b3d51a.png)
which is convex for positive reals 
, we get that ![\[3\sqrt{\frac{3}{ab + bc + ca}} \le \sum_{\mathrm{cyc}} \sqrt{\frac{2}{ac + bc}}.\]](http://latex.artofproblemsolving.com/d/f/a/dfa3dcbd0b9c1770f0273912f6d8d5f94228a34c.png)
By multiplying both sides by 
, this becomes ![\[3\sqrt{3} \cdot \sqrt{\frac{abc}{ab + bc + ca}} \le \sum_{\mathrm{cyc}} \sqrt{\frac{2ab}{a + b}}.\]](http://latex.artofproblemsolving.com/2/2/7/22733d22923a4995db783b347b8798cdf70da52f.png)
Thus, it suffices to show that ![\[\sqrt{\frac{2ab}{a + b}} + \sqrt{\frac{a^2 + b^2}{a + b}} \le \sqrt{2(a + b)}.\]](http://latex.artofproblemsolving.com/b/1/a/b1a5a3f472fba3e8150326a9b8e9991eb5059beb.png)
This follows by Jensen's Inequality on 
, which is concave for positive reals . Thus, we are done.
:
So summing cyclically:
So it suffices to end up showing that 
. This is where we start using the condition. (!)
and 
all in one go, we are going to substitute 
, etc, so the condition reduces down to 
whilst the inequality desired reduces down to
Now notice 
so it suffices to show that
However this is just Jensen with 
or tangent line trick at 
; the actual calculus computation is deferred to desmos (again!) 
![\[
I = \int_{0}^{\frac{1}{\sqrt{3}}} \frac{u \arctan(u)}{(1 - u^2) \sqrt{1 - 2 u^2}} \, du
\]](http://latex.artofproblemsolving.com/b/c/5/bc5d60ff6679870b47776c401658e1ba62af75aa.png)
-
![\[T_1 z = \frac{z + 2}{z + 3}, \qquad T_2 z = \frac z{z + 1},\]](http://latex.artofproblemsolving.com/b/7/2/b72b5e302cecdf2649c43a078aea3aa9a88ef2cc.png)
find 
and 
if and only if 
, then
then the condition becomes 
.
,![\[\sum_{cyc} \sqrt{\frac{z(x^2+y^2)}{(x+y)}} + 3\sqrt{xyz} \le \sum_{cyc} \sqrt{2(x+y)z}\]](http://latex.artofproblemsolving.com/c/0/0/c00f798931b72145b8790ea44c0f6ca5ca5a4e79.png)
Because 
,![\[\sqrt{\frac{z(x^2+y^2)}{(x+y)}} = \sqrt{z\frac{(x+y)^2-2xy}{x+y}} = \sqrt{z(x+y) - \frac{2xyz}{x+y}}\]](http://latex.artofproblemsolving.com/f/f/6/ff6cfdd2215ab2df9a32c9cb2aa01937405ec2eb.png)
By Cauchy we have![\[\sqrt{z(x+y) - \frac{2xyz}{x+y}} + \sqrt{\frac{2xyz}{x+y}} \le \sqrt{2(x+y)z}\]](http://latex.artofproblemsolving.com/7/8/5/785e7fd5c7bc08cce744408b5f1c81e169902f23.png)
So it suffices to have![\[\sum_{cyc} \sqrt{\frac{2}{3-z}} \ge 3\]](http://latex.artofproblemsolving.com/9/9/3/9938a71a747126b715546e7f0e9e336dcedb159e.png)
Because 
By AM-QM: 
Q.E.D.