[/quote]
,
such that there exist infinitely many rational numbers 
satisfying![\[
\left| \frac{q}{p} - \frac{\sqrt{5} - 1}{2} \right| < \frac{\alpha}{p^2}.
\]](http://latex.artofproblemsolving.com/1/1/6/1162c6a7215bff2f9fdaa042a6952f86524e9e92.png)
such that
are all powers of 
.
be a convex quadrilateral. Let 
be the midpoints of 
respectively. If 
concur at a point 
prove that 
,
children standing in a circle. For each girl, it turns out that among the five people following her clockwise, there are more boys than girls. Find the greatest number of girls that can stand in a circle.
so that there exists a unique 
satisfying 
and![\[f(x+g(y))+f(-x-g(-y))=g(x+f(y))+g(-x-f(-y))\]](http://latex.artofproblemsolving.com/d/e/0/de0ea33c3a679ff02762be39e9013b72974e4287.png)
for all 
.
be a triangle with circumcircle 
and incenter 
and let 
be the midpoint of 
.
,
,
are selected on sides 
,
such that 
,
,
.
intersects 
other than 
.
and 
meet on 
interesting if 
is composite and for every divisor 
of 
and 
is also a divisor of 
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Find the minimum possible value of ![\[\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}\]](//latex.artofproblemsolving.com/4/d/2/4d21982ba412c0bb4a7985798e81bc5b01ac0e43.png)
given that 
, 
, 
, 
are nonnegative real numbers such that 
.
Proposed by Titu Andreescu
![\[\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}\]](http://latex.artofproblemsolving.com/4/d/2/4d21982ba412c0bb4a7985798e81bc5b01ac0e43.png)
given that 
, 
, 
, 
are nonnegative real numbers such that 
.Proposed by Titu Andreescu
This post has been edited 3 times. Last edited by djmathman, Jun 22, 2020, 5:50 AM
.
from both sides, we get 
But since 
true for 
,
which is true.
and cyclic variants.![\[ \frac{1}{b^3+4} \ge \frac14 - \frac{b}{12} \]](http://latex.artofproblemsolving.com/b/2/3/b238c461a3ef0839e20f14f8b6d31054b7bab640.png)
since 
.![\[ ab+bc+cd+da = (a+c)(b+d) \le \left( \frac{(a+c)+(b+d)}{2} \right)^2 = 4. \]](http://latex.artofproblemsolving.com/e/a/0/ea0628772ecacef6a9da5ce9d5bdee5892f83ca9.png)
Thus ![\[ \sum_{\text{cyc}} \frac{a}{b^3+4} \ge \frac{a+b+c+d}{4} - \frac{ab+bc+cd+da}{12} \ge 1 - \frac13 = \frac23. \]](http://latex.artofproblemsolving.com/1/f/e/1fe63ef65bd7be5da67cbef0c71c62fda077db39.png)
This minimum 
is achieved at 
and permutations.
is the minimum case...
I also wrote down a short proof for the maximum being 1 cuz I had time left.
achieved at 
.
.
as desired. Now using this tangent-line approximation of the function we find,
Now rearranging the right hand side we find,
as desired.
and 
.
then we have 
and 
is positive for 
and is positive for small 
and has only one root between 
and 
is nonnegative on all nonnegatives. Thus we have ![\[\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4} \ge \frac{a+b+c+d}4-\frac{ab+bc+cd+da}{12}=1-\frac{(a+c)(b+d)}{12}\ge 1-\frac{4}{12}=\frac 23\]](http://latex.artofproblemsolving.com/d/6/2/d6260e0d7221de323ba0ca9c9d1efe154b1d5e84.png)
by AM-GM with equality when 
.
.
to 
,
,
,
.
and permutations. We claim that ![\[\frac{1}{a^3+4} \geq \frac{-a+3}{12}\]](http://latex.artofproblemsolving.com/d/8/f/d8fadd128f30df168cc3b79e3630b55e75db19cc.png)
for ![$a \in [0, 4]$](http://latex.artofproblemsolving.com/c/2/3/c23f3cce1f1f41978d612789e8ca163584de333b.png)
.
we get ![\[b^4-3b^3+4b \geq 0 \Rightarrow b(b+1)(b-2)^2 \geq 0\]](http://latex.artofproblemsolving.com/3/f/8/3f8705768a1f3c56e997c1840165ae286ab4152f.png)
which is obviously true for ![\[\sum_{\text{cyc}} \frac{3a-ab}{12} \geq \frac{2}{3} \Rightarrow \sum_{\text{cyc}} \frac{-ab}{12} \geq \frac{-1}{3} \Rightarrow \sum_{\text{cyc}} ab \leq 4\]](http://latex.artofproblemsolving.com/6/5/0/6508c9a7baab7fe32708854894f084e7694d7b60.png)
This last inequality is true because ![\[\sum_{\text{cyc}} ab = (a+c)(b+d) = (2-x)(2+x) \leq 4\]](http://latex.artofproblemsolving.com/5/a/9/5a98ba030ed6fc943deb876c10c8927fb196d1db.png)
.![\[\frac{1}{b^3+4} \geq \frac{1}{4} - \frac{b}{12}.\]](http://latex.artofproblemsolving.com/4/1/2/412925a2200b5b8327dd00052ad8601c8785b4e6.png)
![\[(a+c)(b+d) \leq \left(\frac{a+b+c+d}{2}\right)^2 = 4.\]](http://latex.artofproblemsolving.com/9/5/9/9596a00af2884adf224aebdf426f1ea6e6b60130.png)
![\[\sum_{cyc}\frac{a}{b^3+4} \geq \frac{a+b+c+d}{4} - \frac{ab+bc+cd+da}{12} \geq 1 - \frac{1}{3} = \frac{2}{3}.\]](http://latex.artofproblemsolving.com/d/6/d/d6dee1f2b5c2d7f2cadbfc8c7b6878e03b78e9bd.png)
for nonnegative 
with equality at 
Next note that 
Therefore the original expression is
which is achievable at 
Equality holds when 
