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.
, where a > 0 and 
is differentiable such that 
Then which of the following is correct?
always exists but not necessarily zero. always exists and is equal to zero. may not exist. exists if 
is twice differentiable.
such that for all 
, ![\[xf(y+x)+(y+x)f(y)=f(x^2+y^2)+2f(xy)\]](http://latex.artofproblemsolving.com/5/b/1/5b101ead6195155ccbd99b757016e8f73dac4411.png)
Proposed by Aritra Mondal
be positive real numbers such that 
. Prove that
with rational coordinates lie on a plane. It turned out that the distance between every pair of points is an integer. Prove that there exist points 
with integer coordinates such that 
, 
, 
checkered board is cut into several squares with integer side length. The size of a square is it's side length.
of integers, a pair 
with 
is called interesting if there exists a pair 
of integers with 
such that 
For each 
, find the largest possible number of interesting pairs in a sequence of length 
.
such that, for all 
,
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 .
be positive real number such that 
Find the minimum value of 
and 
equals one, then 
not 
But with the same substitutions, we can use alternative C-S solution (and Vasile's inequality) like below:
not But with the same substitutions, we can use alternative C-S solution (and Vasile's inequality) like below:
and be positive numbers . Prove that 
be positive numbers . Prove that 
and be positive numbers . Prove that 
, the Cauchy-Schwarz's Inequality gives that
It is easy to show that 
. Therefore, the inequality ![$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{3}{\sqrt[3]{abc}}$](http://latex.artofproblemsolving.com/f/4/1/f41b3d1ee10a3656bbae0ce631a1c38b24d6b96c.png)
implies![\begin{align*}\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge\frac{\frac{9}{\sqrt[3]{(abc)^2}}}{3+\frac{3}{\sqrt[3]{abc}}}=\frac{3}{\sqrt[3]{abc}(\sqrt[3]{abc}+1)}\ge\frac{3}{1+abc},
\end{align*}](http://latex.artofproblemsolving.com/8/4/b/84b96b1dabde2948d23d6ab6fcd9e2ac53f94973.png)
where the last inequality holds since 
. We are done. 
. Prove that
(Lijvzhi)
Prove that![\[ \frac 1{a(2+a)}+\frac 1{b(2+b)}+\frac 1{c(2+c)}+\frac 1{d(2+d)}\geq \frac 4{ 2+abcd}\]](http://latex.artofproblemsolving.com/0/4/d/04d223126d59422752a97e762b601cc298191683.png)
Solution of ytChen 
. Using the Cauchy-Schwarz's Inequality and AM-GM Inequality, we obtain![\begin{align*}\frac 1{a(2+a)}+\frac 1{b(2+b)}+\frac 1{c(2+c)}+\frac 1{d(2+d)}=&\frac{x^2}{2x+1}+\frac{y^2}{2y+1}+\frac{z^2}{2z+1}+\frac{u^2}{2u+1}\\
\ge&\frac{(x+y+z+u)^2}{2(x+y+z+u)+4}\\
=&\frac{1}{\frac{2}{x+y+z+u}+\frac{4}{(x+y+z+u)^2}}\\
\ge&\frac{1}{\frac{1}{2\sqrt[4]{xyzu}}+\frac{1}{4\sqrt[4]{(xyzu)^2}}}\\
=&\frac{4}{2\sqrt[4]{abcd}+\sqrt[4]{(abcd)^2}}\\
\ge&\frac{4}{2+abcd},
\end{align*}](http://latex.artofproblemsolving.com/3/e/5/3e5b9b4d313d736f137b85c77060bd4cc4455000.png)
where the last inequality holds since ![$w^4+2-w^2-2w=(w-1)^2(w^2+2w+2)\ge0\ \left(\forall w=\sqrt[4]{abcd}\right)$](http://latex.artofproblemsolving.com/d/3/b/d3b87e901d52d2a6b3ba7dd5104500674c68b4ed.png)
. We are done.
. Prove that(Lijvzhi)
. Using the C-S and AM-GM , we obtain![\begin{align*}\frac{1}{a_1\left(a_1+n-2\right)}+\frac{1}{a_2\left(a_2+n-2\right)}+\cdots+\frac{1}{a_n\left(a_n+n-2\right)}\\=&\frac{b_1^2}{(n-2)b_1+1}+\frac{b_2^2}{(n-2)b_2+1}+\cdots+\frac{b_n^2}{(n-2)b_n+1}\\
\ge&\frac{(b_1+b_2+\cdots+b_n)^2}{(n-2)(b_1+b_2+\cdots+b_n)+n}\\
=&\frac{1}{\frac{n-2}{b_1+b_2+\cdots+b_n}+\frac{n}{(b_1+b_2+\cdots+b_n)^2}}\\
\ge&\frac{1}{\frac{n-2}{n\sqrt[n]{b_1b_2\cdots b_n}}+\frac{1}{n\sqrt[n]{(b_1b_2\cdots b_n)^2}}}\\
=&\frac{n}{(n-2)\sqrt[n]{a_1a_2\cdots a_n}+\sqrt[n]{(a_1a_2\cdots a_n)^2}}\\
\ge&\frac{n}{n-2+a_1a_2\cdots a_n},
\end{align*}](http://latex.artofproblemsolving.com/8/1/2/81201ba705423fef3914cd56b28fac342bfc815a.png)
where the last inequality holds since 
![$\forall w=\sqrt[n]{a_1a_2\cdots a_n}$](http://latex.artofproblemsolving.com/a/b/f/abf9b837da0fb966837e253cac8a8ebe7fc0e365.png)
. We are done. 
prove inequality 
Prove that
Where 
. Prove that
Let 
. Prove that
. Prove that