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such that 
is an integer.
cards on which the numbers 
, 
, 
, are written respectively. Alice and Bob play the following game: in each turn, Alice gives two cards to Bob, who must keep one card and discard the other. The game proceeds for four turns in total; in the first two turns, Bob cannot keep both of the cards with the larger numbers, and in the last two turns, Bob also cannot keep both of the cards with the larger numbers. Let 
be the sum of the numbers written on the cards that Bob keeps. Find the greatest positive integer 
for which Bob can guarantee that is at least .
such that
for any positive integers 
, 
.
, with a,b,c,d are arbitrary integers and p is an arbitrary prime such that the fraction is an integer too.
and 
positive real numbers. Prove that![\[\frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+\cdots+\frac{1+x_{n}^2}{1+x_{n}x_{1}}\geq n.\]](http://latex.artofproblemsolving.com/e/9/e/e9e73a14f4dfa6315e7562998a9394d371847139.png)
and 
Prove that
Let 
and Prove that
Let 
and Prove that
Let 
and 
Show that![\[\sum_{k=1}^n\frac{1}{1+\sum_{j\neq k}x_j}\le\frac n{1+(n-1)\sqrt[n]{x_1x_2\cdots x_n}}.\]](http://latex.artofproblemsolving.com/e/4/c/e4ca07e5003fad8937af3c44dcbc9906fa48e5b4.png)
and 
. Prove that
.
, then![\[
a+b+a^2+b^2 \geq 1 \geq 28-6\sqrt{21}.
\]](http://latex.artofproblemsolving.com/6/7/7/677ab9d63defffddb7a8c986c2e2b4ddb4e8267e.png)
, then![\[
\frac{1}{b^2-b+1} \leq \frac{1}{a^2-a+1}, a^2 + a + 1 \geq b^2 + b + 1
\]](http://latex.artofproblemsolving.com/e/8/f/e8fdd29170a984431f76f172d52feb9724f7edfc.png)
which implies that![\[
\frac{a^2+a+1}{a^2-a+1}+\frac{b^2+b+1}{b^2-b+1} \geq \frac{a^2+a+1}{b^2-b+1}+\frac{b^2+b+1}{a^2-a+1} \geq 3.
\]](http://latex.artofproblemsolving.com/8/a/3/8a3d11ba00dc8246113ff3c5881989a12be3f4c9.png)
![\[
3 \geq 6-\left(\frac{a^2+a+1}{a^2-a+1}+\frac{b^2+b+1}{b^2-b+1}\right)
=\frac{2(a-1)^2}{a^2-a+1}+\frac{2(b-1)^2}{b^2-b+1}
\geq \frac{2(1-a+1-b)^2}{a^2-a+1+b^2-b+1}.
\]](http://latex.artofproblemsolving.com/2/2/2/222d49414291c3fd180471f92d561d4d0f69b171.png)
and 
. Then,![\[
3\left(u^2-u+1-2v\right) \geq 2\,(2-u)^2.
\]](http://latex.artofproblemsolving.com/2/a/4/2a4022474a7acca1c57c06b05a98c40e269e5c80.png)
![\[
u^2-4v \geq \frac{u^2-10u+4}{3}.
\]](http://latex.artofproblemsolving.com/c/2/4/c24df229451c119fb2ac4cac1352030302fa6f9a.png)
![\[
a+b+a^2+b^2 = u^2-2v+u = \frac{u^2-4v}{2}+\frac{u^2}{2}+u,
\]](http://latex.artofproblemsolving.com/4/9/4/49466166b20f3199205c898f5de4e53481919827.png)
if 
, then![\[
\frac{u^2-4v}{2}+\frac{u^2}{2}+u \geq \frac{u^2-10u+4}{6}+\frac{u^2}{2}+u
=\frac{2}{3}\,(u^2-u+1).
\]](http://latex.artofproblemsolving.com/d/7/2/d72ef57efa4ba4f14b344a02649dd4c961867fd7.png)
is monotonically decreasing on ![\( u\in \left(0,5-\sqrt{21}\right] \)](http://latex.artofproblemsolving.com/2/b/d/2bde58e2769d4081e0740823e379e1ed883a7ca5.png)
, it follows that![\[
\frac{2}{3}\,(u^2-u+1) \geq \frac{2}{3}\Bigl[(5-\sqrt{21})^2-(5-\sqrt{21})+1\Bigr]
= 28-6\sqrt{21}.
\]](http://latex.artofproblemsolving.com/d/b/c/dbc2c7413768ebff082b089c4c28ffa2197d6a03.png)
, then since 
, we have![\[
\frac{u^2-4v}{2}+\frac{u^2}{2}+u \geq \frac{u^2}{2}+u \geq \frac{(5-\sqrt{21})^2}{2}+(5-\sqrt{21})
= 28-6\sqrt{21}.
\]](http://latex.artofproblemsolving.com/b/4/7/b4785f87dac778351daf902c356f8c2409418183.png)
![\[
a+b+a^2+b^2 \geq 28-6\sqrt{21}.
\]](http://latex.artofproblemsolving.com/2/8/1/28126ac3dc744ba5841d2065f3965fb53db64d91.png)
such that 
Prove that
Prove that
.
Let 
.
Let 
.
.
Let 
.
then it suffices to prove 
and 
hence 
with equality at 
and 
Show that
Let 
Show that
Let 
Show that
Let 
Show that
Let 
be reals such that 
Show that
Let 
Show that