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 positive real numbers. Prove the inequality![\[
\frac{x_1 + 3x_2}{x_2 + x_3}
\;+\;
\frac{x_2 + 3x_3}{x_3 + x_4}
\;+\;
\frac{x_3 + 3x_4}{x_4 + x_1}
\;+\;
\frac{x_4 + 3x_1}{x_1 + x_2}
\;\ge\;8.
\]](http://latex.artofproblemsolving.com/5/c/0/5c062a37f6ff120232ae87eb0093a4ce73ee0eb1.png)
+1 w
be a function satisfying the following conditions for all 
:![\[
\begin{cases}
f(n+1) > f(n) \\
f(f(n)) = 2n + 2
\end{cases}
\]](http://latex.artofproblemsolving.com/0/4/7/047e28c9d899d7de443ae40ac2c756fda851fafb.png)
Find the value of 
.
is the foot of the altitude from 
of triangle 
. On the lines 
and 
points 
and 
are marked such that the circumcircles of triangles 
and 
are tangent, call this circles 
and 
respectively. Tangent lines to circles and at and intersect at 
.
.
be a triangle with side‐lengths 
, incenter 
, and circumradius 
. Denote by 
the area of , and let 
be the areas of triangles 
, 
, and 
, respectively. Prove that![\[
\frac{abc}{12R}
\;\le\;
\frac{P_1^2 + P_2^2 + P_3^2}{P}
\;\le\;
\frac{3R^3}{4\sqrt[3]{abc}}.
\]](http://latex.artofproblemsolving.com/1/3/9/139a3e15f08afb78ca0b8301f24d30db8f202207.png)
be an acute scalene triangle. Denote by 
the circle with diameter 
, and let 
be the contact points of the tangents from to , chosen so that 
and 
lie on opposite sides of and 
and 
lie on opposite sides of . Similarly, let 
be the circle with diameter 
, with tangents from touching at 
, and 
the circle with diameter , with tangents from touching at 
.
are concurrent.
and 
are prime and satisfy![\[\frac{p}{{p + 1}} + \frac{{q + 1}}{q} = \frac{{2n}}{{n + 2}}\]](http://latex.artofproblemsolving.com/7/3/5/735ec9416a654f40121021bdd1233ffa86e42495.png)
. Find all possible values of 
.
with orthocenter 
and a point 
lying on the Euler line of 
Prove that the midpoint of 
lies on the Thomson cubic of the pedal triangle of WRT
be an acute triangle with 
, and let 
be the center of its circumcircle. Let 
be the reflection of 
with respect to 
. The line through parallel to intersects 
at 
, and the tangent at to the circle 
intersects the line through parallel to at point 
. Let 
be a point on the ray 
, starting at , such that 
. lies on the circle with diameter 
.
is written on the board, with 
linear factors on each side. What is the least possible value of 
for which it is possible to erase exactly of these 
linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?
![\[\frac{a}{2(a^2-bc)+1}+\frac{b}{2(b^2-ca)+1}+\frac{c}{2(c^2-ab)+1}\leq \frac{1}{a+b+c}\]](http://latex.artofproblemsolving.com/1/3/6/136a1e915288b7988899ecd1b045676acaf0735b.png)
the inequality can be written as ![\[ \sum \frac{a(a+b+c)}{2a(a+b+c)+(ab+bc+ca)} \le 1,\]](http://latex.artofproblemsolving.com/f/b/6/fb62f3b6963ec701c5efb75b4248b783091a9573.png)
or ![\[\sum \frac{2a(a+b+c)}{2a(a+b+c)+(ab+bc+ca)} \le 2.\]](http://latex.artofproblemsolving.com/3/9/5/39518ea316b5b36a2d462c8751726839098dac27.png)
Now, notice that ![\[\frac{2a(a+b+c)}{2a(a+b+c)+(ab+bc+ca)}=1-\frac{(ab+bc+ca)}{2a(a+b+c)+(ab+bc+ca)}.\]](http://latex.artofproblemsolving.com/4/0/7/407ac70557dd792fdf258e122a7534d0aead96e0.png)
Therefore, the inequality is equivalent to ![\[(ab+bc+ca) \sum \frac{1}{2a(a+b+c)+(ab+bc+ca)} \ge 1.\]](http://latex.artofproblemsolving.com/4/5/8/4587b3e84c5a2c46bb8cb446a8c4c75deb808174.png)
Using the Cauchy-Schwarz inequality, we get ![\[\begin{aligned} \sum \frac{1}{2a(a+b+c)+(ab+bc+ca)} & \ge \frac{\left(\sum bc\right)^2}{\sum b^2c^2\left[2a(a+b+c)+(ab+bc+ca)\right]} \\ &=\frac{\left(\sum bc\right)^2}{2abc(a+b+c)\sum bc +(ab+bc+ca)\sum b^2c^2} \\ &=\frac{1}{ab+bc+ca}. \end{aligned} \]](http://latex.artofproblemsolving.com/7/f/8/7f81c6a5f1bc156bc353e7e90da6aeef7b0a0eb6.png)
From this estimation, the conclusion follows. We are done. 
From this estimation, the conclusion follows. We are done. Could you tell me how you thought of that please?! 
be positive real numbers for which 
. Prove the inequality![\[ 1\geq \frac{a}{a^2-bc+1}+\frac{b}{b^2-ca+1}+\frac{c}{c^2-ab+1}\ge\frac{1}{a+b+c}\]](http://latex.artofproblemsolving.com/a/f/c/afc7e8bfcec3ac1ffe5771eaf2f28eda8d63ecf0.png)
.
.![\[\sum ab +\sum ab^2 \le 6\]](http://latex.artofproblemsolving.com/9/f/6/9f632ab1e6df5029bcd4416eb06623123ad2cd8c.png)
P![$$ \frac{1}{x^2} + 2y^2 + 3x + \frac{4}{y} \geq 3\left(2+\sqrt[3]{\frac{9}{4} }\right)$$](http://latex.artofproblemsolving.com/6/b/6/6b6011a32b103fe94652974a521eeae479cba1d8.png)
.
is the inverse of function 
and thus, they must intersect that 
axis (as its clearly not self-inverse function). Thus: 
is the only solution.
. Prove that
W here 
![\[ \frac{a}{a^2-bc+1}+\frac{b}{b^2-ca+1}+\frac{c}{c^2-ab+1}\ge\frac{1}{a+b+c}\]](http://latex.artofproblemsolving.com/9/2/f/92fc4af06826d9b973ad96bca902f32d17f71a6e.png)
![\[ \frac{a}{a^2-bc+1} + \frac{b}{b^2-ca+1} + \frac{c}{c^2-ab+1} \ge \frac{(a+b+c)^2}{a^3+b^3+c^3-3abc+a+b+c}= \frac{1}{a+b+c} \]](http://latex.artofproblemsolving.com/3/5/1/3514e84ca23ac4333f9e51764f305f7facde2a57.png)
![\[\sum{\frac{a}{a^2-bc+1}}-\frac{1}{a+b+c}=\sum{\frac{ab(3ac+3bc+3c^2+2)(a-b)^2}{3(a+b+c)(a^2-bc+1)(b^2-ca+1)(c^2-ab+1)}}\]](http://latex.artofproblemsolving.com/3/9/4/394be8b2163df539c536973abbd21dbc7ed4327a.png)
![\[ \frac{1}{a^2-bc+1}+\frac{1}{b^2-ca+1}+\frac{1}{c^2-ab+1}\le3\]](http://latex.artofproblemsolving.com/f/8/9/f89475a6d140dc5ebf7f9e6ffa53fa0839d4d41d.png)