![$Q\in \{-1,1,0\}[x]$](http://latex.artofproblemsolving.com/b/4/0/b404c8d6415aaf3d4442cc5381f582c1f8f8b996.png)
with 
.![$\exists P \in \mathbb Z[x]$](http://latex.artofproblemsolving.com/a/2/e/a2e67dd74df9971ca90d1a5d81c1c8e1f3982276.png)
palindrom with 
?
for which 
,
denote the integer part.
such that 
is divisible by 
,
is divisible by 
,
is divisible by 
.
and 
Prove that
Y by Adventure10, chessgocube, Mango247
Let 
, 
, 
be three positive reals such that 
. Prove that the following inequality holds: ![\[ ab+bc+ca \leq \frac 34 . \]](//latex.artofproblemsolving.com/8/0/2/8022b2c879b43fd65e34cf48b4d1a93ea38fac8a.png)
Cezar Lupu
, 
, 
be three positive reals such that 
. Prove that the following inequality holds: ![\[ ab+bc+ca \leq \frac 34 . \]](http://latex.artofproblemsolving.com/8/0/2/8022b2c879b43fd65e34cf48b4d1a93ea38fac8a.png)
Cezar Lupu
.
.
.
for some real k > 1. Hence, if we define 
,
and 
,
.
,
,![$\frac{BC+CA+AB}{3}\geq \sqrt[3]{BC\cdot CA\cdot AB}$](http://latex.artofproblemsolving.com/f/5/c/f5c154f068376a3a08df38c4809b2ce2cc70df20.png)
.
,![$\frac{\left(\frac34\right)}{3}\geq \sqrt[3]{BC\cdot CA\cdot AB}$](http://latex.artofproblemsolving.com/e/3/7/e372a8fffc14825faeb141edda6623711088cb36.png)
;![$\frac14\geq \sqrt[3]{A^2B^2C^2}$](http://latex.artofproblemsolving.com/2/b/d/2bdd14ba1cd38bd86412945234501b823414260b.png)
.![$\frac12\geq\sqrt[3]{ABC}$](http://latex.artofproblemsolving.com/9/e/b/9eb6ddc6a8529272dc03d261911825fa0c9d1b6c.png)
;
.
.
;
.
.
.
.
![\[
1 = (a + b)(b + c)(c + a) \geqslant \frac{8}
{9}(a + b + c)(ab + bc + ca) \geqslant \frac{8}
{9}(ab + bc + ca)\sqrt {3\left( {ab + bc + ca} \right)}
\]](http://latex.artofproblemsolving.com/2/0/c/20c0055eacaf25b81b60bda1885b08f35de19438.png)
![\[
\left( {ab + bc + ca} \right)^{\frac{3}
{2}} \leqslant \frac{{3\sqrt 3 }}
{8} \Rightarrow ab + bc + ca \leqslant \frac{3}
{4}
\]](http://latex.artofproblemsolving.com/9/4/d/94df7d225bda2b0012b6671c4df42340cdbf1756.png)
.
by one application of Schur and AM-GM, we see
and use the well-known inequality 
be positive reals such that 
.
is the unique positive real solution to 
,
and 
is a counterexample. (found with computer)
and 
and we have:![\[ab+bc+ac \leq \frac{2}{3}(a+b+c)(ab+bc+ca)=\frac{2}{3}(1+abc) \leq \frac{2}{3}(1+\frac{1}{8})=\frac{3}{4}\]](http://latex.artofproblemsolving.com/7/b/4/7b4700eb56377444045f73ba3c4ada1478cc34a0.png)
Homogenize and get a trivial inequality, even stronger than the quoted one
But even so, it's much stronger than the starting inequality
and 
and ![$(a+b) + (b+c) + (c+a) \ge 3\sqrt[3]{(a+b)(b+c)(c+a)} \implies a+b+c \ge \frac{3}{2}$](http://latex.artofproblemsolving.com/8/6/1/8614b8b20fb4901dd0b1f656db4e9d055fa4529a.png)
Now Note that we need to prove 
which now is obvious.
