Let $x,y,z$ be positive real numbers satisfying $x+y+z=1$. Find the smallest $k$ such that $\frac{x^2y^2}{1-z}+\frac{y^2z^2}{1-x}+\frac{z^2x^2}{1-y}\leq k-3xyz$.
Problem
Source: 2016 Taiwan TST Round 3
Tags: inequalities
Grotex
23.07.2016 17:23
$k=\frac{1}{6}.$
Diamondhead
23.07.2016 17:38
The answer is $\frac{1}{6}$. Note that if we put $x=y=z=\frac{1}{3}$, then $k \geq \frac{1}{6}$.
We will prove that $\frac{x^2y^2}{1-z}+\frac{y^2z^2}{1-x}+\frac{z^2x^2}{1-y}\leq \frac{1}{6}-3xyz$.
From A.M-G.M inequality, $2(x^3+y^3+z^3)+3(x^2y+xy^2+y^2z+yz^2+z^2x+zx^2) \geq 24xyz$.
Therefore, $\frac{1}{4}(x^2y+xy^2+y^2z+yz^2+z^2x+zx^2)+3xyz \leq \frac{1}{6}((x^3+y^3+z^3)+3(x^2y+xy^2+y^2z+yz^2+z^2x+zx^2)+6xyz)$
So, $\frac{1}{4}(x^2y+xy^2+y^2z+yz^2+z^2x+zx^2)+3xyz \leq \frac{1}{6}(x+y+z)^3=\frac{1}{6}.$
Therefore, $\frac{1}{4}xy(x+y)+\frac{1}{4}yz(y+z)+\frac{1}{4}zx(z+x)+3xyz \leq \frac{1}{6}.$
Since $\frac{1}{4}xy(x+y) \geq \frac{x^2y^2}{x+y}$, $\frac{1}{4}yz(y+z) \geq \frac{y^2z^2}{y+z}$, and $\frac{1}{4}zx(z+x) \geq \frac{z^2x^2}{z+x}$, therefore, $$\frac{x^2y^2}{x+y}+\frac{y^2z^2}{y+z}+\frac{z^2x^2}{z+x}+3xyz \leq \frac{1}{6}.$$So, $\frac{x^2y^2}{1-z}+\frac{y^2z^2}{1-x}+\frac{z^2x^2}{1-y}\leq \frac{1}{6}-3xyz$ as desired.
MonsterS
10.03.2017 15:38
$k = 6$ $ \sum{\frac{x^2y^2}{x+y}} \leq \sum{\frac{xy(x+y)}{4}}$ by AM-HM So we need to prove. $9xyz+(\sum{xy})(\sum{x}) \leq \frac{2}{3}$ by AM-GM$ \implies 9xyz+(\sum{xy})(\sum{x}) \leq \frac{(\sum{x})^3}{3}+\frac{(\sum{x})^3}{3}=\frac{2}{3}$
bin_sherlo
18.02.2024 16:26
$k=\frac{1}{6}$ holds when $x=y=z=\frac{1}{3}$ \[\sum{(\frac{x^2y^2}{x+y}+xyz)}=\sum{xy(\frac{xy+yz+zx}{x+y})}=\sum{xy}.\sum{\frac{xy}{x+y}}\leq \sum{xy}.\sum{\frac{\sqrt{xy}}{2}}\]\[\sum{xy}.\sum{\frac{\sqrt{xy}}{2}}\leq \frac{1}{2}.\sum{xy}.\sqrt{\sum{xy}.3}\leq \frac{1}{2}.\frac{(\sum{x})^2}{3}\sqrt{\frac{(\sum{x})^2}{3}.3}=\frac{1}{6}\blacksquare\]