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.
written on the board.At every step,Mikail chooses the two numbers 
and substitutes them with 
and 
where 
is the unchosen number on the board. Prove that at least 
negative number must be remained on the board at any step.
satisfying the conditions:
.
![$f(x)\in \mathbb{C} [x]$](http://latex.artofproblemsolving.com/d/c/8/dc8a579c6583cd51d891239ea862f93bfe740846.png)
, s.t. 
be positive real numbers. such that![\[47(a^2+b^2+c^2+d^2) - 5(a+b+c+d)^2 = 27.\]](http://latex.artofproblemsolving.com/5/b/e/5be6b2c96d5761d3a1a257b8bf3cdfc15c3e4db5.png)
Prove that![\[\frac{a^2 + b^2 + c^2}{a + b + c}+\frac{b^2 + c^2 + d^2}{b + c + d}+ \frac{c^2 + d^2+a^2}{c + d+a}+ \frac{d^2+a^2 + b^2}{d+a + b} \geqslant 2.\]](http://latex.artofproblemsolving.com/3/c/8/3c8db9e517e825e311129255180672658cff90cb.png)
be positive real numbers. Prove that![\[\sum_{i=1}^{n} \frac{\displaystyle\sum_{j=1}^{n} a_j^2 - a_i^2}{\displaystyle \sum_{j=1}^{n} a_j - a_i} \geqslant \sqrt{\displaystyle \frac{n(2n^3 - 6n^2 + 4n - 1)}{(n - 1)^3}\sum_{j=1}^{n} a_j^2
- \frac{n(n^2 - 3n + 1)}{(n - 1)^3}\left(\sum_{i=1}^{n} a_i \right)^2}.\]](http://latex.artofproblemsolving.com/b/a/c/bac25e978624c8ea5c82c1a4975bb5973fe00135.png)
. The incircle of triangle is tangent to the sides 
and 
at points 
and 
, respectively. Let 
and 
be the midpoints of 
and 
, respectively. Let 
and 
be the midpoints of 
and 
, respectively. Prove that 
and 
intersect on the perpendicular bisector of side 
.
be a triangle with circumcenter 
.
is a fix line passing through and 
an arbitrary point on it . denote 
the circumcircle of pedal triangle of . prove that passese through a fix point
be reals such that 
. Prove that
T729. Let 
. Prove that
T730. Let . Prove that
such that ![$\sqrt[3]{1}=x$](http://latex.artofproblemsolving.com/e/d/9/ed965489f2f983dbb4669c9e94ea9aa83baa72a7.png)
.
such that ![$\sqrt[4]{1}=y$](http://latex.artofproblemsolving.com/d/1/6/d16559d02217f335c54309bd3effb04e6e2f4b01.png)
.
such that ![$\sqrt[5]{1}=z$](http://latex.artofproblemsolving.com/3/3/d/33d87611165181f89f950addf1afb749c1b01df9.png)
.
.
is always ; however, if you write 
, you would get the correct solution sets.
is always ; however, if you write , you would get the correct solution sets.
is always ; however, if you write , you would get the correct solution sets.
is always ; however, if you write , you would get the correct solution sets.![$\sqrt[3]{x}$](http://latex.artofproblemsolving.com/d/d/2/dd20cd4b23d9242a4349d63b1de7514bc295f39c.png)
is a function.
is always ; however, if you write , you would get the correct solution sets. is a function.![$f(x) = \sqrt[3]{x}$](http://latex.artofproblemsolving.com/e/d/a/eda98cca3591121f041ec3d26a4aad8cc82dfe78.png)
is a function, and if 
(the number who when cubed is equal to the original number) then is a cube root of 
so he is correct.
is always ; however, if you write , you would get the correct solution sets. is a function.![$\sqrt[3]{1}=x\iff x^3=1,$](http://latex.artofproblemsolving.com/5/8/c/58c31e22cf92cba6a1d8c62bf2c9da62999ec784.png)
which has 
solutions, which can be found via roots of unity
Prove that
Prove that
Let 
Prove that
Let 
Prove that
Prove that
or 
![$P(x)\in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/4/0/9/4099817b1a673007a2b2ba0f5ebc6a0697eadcad.png)
such that 
Prove that
Let 
Prove that
Prove that
Prove that
Where 
