[/quote]
,
cards on which the numbers 
,
,
,
be the sum of the numbers written on the cards that Bob keeps. Find the greatest positive integer 
for which Bob can guarantee that 
such that
,
.
, with a,b,c,d are arbitrary integers and p is an arbitrary prime such that the fraction is an integer too.
Prove that
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 
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)
with 
the following inequality holds:
,
,
,![\[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc}
\]](http://latex.artofproblemsolving.com/f/a/9/fa964e29de91f0e2f3cb32e335a52d4521bc38fd.png)
.![\[ \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}.
\]](http://latex.artofproblemsolving.com/b/2/9/b295b0b2a765e41a639657e13b724c60323fa208.png)
where 
is in radians.
be distinct reals such that 
Prove that 
be the altitude and 
be the intersection point of the altitudes of triangle 
.
.
intersects 
.
.
Y by
It is known that the moving point 
is on the curve 
. There is a parabola 
with focus 
. Two tangent lines to
are drawn through a point 
on 
, and the tangent points are 
and 
respectively. The line 
parallel to the line 
is tangent to at point 
. Question: When the line and 
have two intersection points, find the range of 
.\
This may make some of the information in the question useless, because I deleted the first two questions of this big question in order to avoid making the question too long and get straight to the point.\
According to the calculation, I get the analytical expression of the line as 
.\
At first, I thought it only needed to be not parallel to the parabola asymptotes.\
That is, the slope of the straight line
, then 
, so 
,and 
.\
.\
But when I checked the answer, it was wrong.\
It combines the straight line with the curve to get an equation and then uses Vieta's theorem.
\begin{cases}
y=\frac{x_0}{2} x - \frac{x_0^2}{4}
x^2=4y
\end{cases}
The answer is that according to the question, the equation has two negative roots. I can't understand this, and this is exactly where my problem lies.\
Then it gets the following system of equations:
\begin{cases}
4x_0-16 \neq 0
\Delta > 0
x_1+x_2 <0
x_1\cdot x_2>0
\end{cases}
Solve,
.
So we get
.
I hope you can help me figure out why both roots of that equation are negative.
is on the curve 
. There is a parabola 
with focus 
. Two tangent lines to
are drawn through a point 
on 
, and the tangent points are 
and 
respectively. The line 
parallel to the line 
is tangent to at point 
. Question: When the line and 
have two intersection points, find the range of 
.\This may make some of the information in the question useless, because I deleted the first two questions of this big question in order to avoid making the question too long and get straight to the point.\
According to the calculation, I get the analytical expression of the line
as 
.\At first, I thought it only needed to be not parallel to the parabola asymptotes.\
That is, the slope of the straight line
, then 
, so 
,and 
.\
.\But when I checked the answer, it was wrong.\
It combines the straight line
with the curve to get an equation and then uses Vieta's theorem.\begin{cases}
y=\frac{x_0}{2} x - \frac{x_0^2}{4}
x^2=4y
\end{cases}
The answer is that according to the question, the equation has two negative roots. I can't understand this, and this is exactly where my problem lies.\Then it gets the following system of equations:
\begin{cases}
4x_0-16 \neq 0
\Delta > 0
x_1+x_2 <0
x_1\cdot x_2>0
\end{cases}
Solve,
.So we get
.I hope you can help me figure out why both roots of that equation are negative.
and 
,
