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+2 w![\[ \lfloor u \cdot \sqrt{5} + v \cdot \sqrt{7} \rfloor \]](http://latex.artofproblemsolving.com/5/3/d/53d029218ac9fac36be38a755ad59a8b4a9b6f8f.png)
where u and v are positive integers.
+1 w
be an acute-angled triangle with 
, let 
be its circumcentre, and let 
be a point on the segment 
. The line through perpendicular to intersects the lines 
and 
at 
and 
respectively. The circumcircles of triangles 
and intersect again at 
.
and 
then 
is tangent to the circle 
has side lengths 
, 
, and 
. Select a point 
inside , and construct the incenters of 
, 
, and 
and denote them as 
, 
, 
. What is the maximum area of the triangle 
?
such that 
. Prove that 
May everyone not to upload solutions on this problem anymore until May 15/2025, this is an active problem on Mathematical Reflection! (Thank you Victoria_Discalceata1)
Hence, it's enough to prove 
then
Equality cases: 
such that . Prove that 
then 
We have![\[a^3b+b^3c+c^3a \leqslant a^3b+b^3c+c^3a + c(b+c)(a-b)(b-c) = b(c^3+abc+a^3).\]](http://latex.artofproblemsolving.com/a/3/b/a3b661208070cac04b6174c698638c1c5e427328.png)
Thus![\[\begin{aligned}
a^3b+b^3c+c^3a+9abc &\leqslant b(c^3+abc+a^3)+9abc \\
& = b[(c+a)^3-3ca(c+a)+(b+9)ca] \\
& = b[(3-b)^3-3ca(3-b)+(b+9)ca] \\
& = b[4bca-(b-3)^3] \\
& \leqslant b[b(c+a)^2-(b-3)^3] \\
& \leqslant 3b(b-3)^2.
\end{aligned}\]](http://latex.artofproblemsolving.com/9/8/8/988d5dd48718e002f93bdfa7ff8b211abd3c6055.png)
It remains to prove that![\[b(b-3)^2 \leqslant 4,\]](http://latex.artofproblemsolving.com/c/b/9/cb95af570bf9666e4c9d5213e55843b2c7f4dfc4.png)
or![\[(4-b)(b-1)^2 \geqslant 0.\]](http://latex.artofproblemsolving.com/f/c/7/fc77075fa28a0e371a000627785880fe6256de30.png)
Which is true.
, it suffices to show that![\[4(a+b+c)^3\ge27\left(a^3b+b^3c+c^3a\right)+81abc(a+b+c),\tag{1}\]](http://latex.artofproblemsolving.com/2/9/4/29424447811d740c1290746604007e21cbd3496a.png)
or![\[\sum_{\rm cyc}a^4+6\sum_{\rm cyc}a^2b^2-\frac{33}4\sum_{\rm cyc}a^2bc\ge\frac{11}4\sum_{\rm cyc}a^3b-4\sum_{\rm cyc}a^3c.\tag{2}\]](http://latex.artofproblemsolving.com/0/3/8/038fccd2b9294dab8674f08fa9f184b5eade1c7d.png)
![\[\sum_{\rm cyc}a^4+r\sum_{\rm cyc}a^2b^2+(p+q-r-1)\sum_{\rm cyc}a^2bc\ge p\sum_{\rm cyc}a^3b+q\sum_{\rm cyc}a^3c\]](http://latex.artofproblemsolving.com/4/5/3/453c0ef90919246a1d442fac357fa944cb1f281e.png)
holds for all real numbers , , if and only if 
. Since![\[3(1+6)=21\ge\left(\frac{11}4\right)^2+(-4)^2+\frac{11}4(-4)=\frac{201}{16},\]](http://latex.artofproblemsolving.com/6/4/a/64a8c14171fbab32f7d92c9f12f0bf9db93c3ad7.png)
then (2) indeed holds. We are done.![\[\frac18\sum_{\rm cyc}\left(4 a^2-2ab+9ac-7bc-4c^2\right)^2+\frac{45}8\sum_{\rm cyc}a^2(b-c)^2\ge0.\]](http://latex.artofproblemsolving.com/2/0/6/206d2dbd6999a8962e7610d31505505214234271.png)
holds for all 
with 
and 
Equality holds when 
or 
or 
