Difference between revisions of "2017 JBMO Problems/Problem 2"

Revision as of 09:25, 24 June 2019

Let $x,y,z$ be positive integers such that $x\neq y\neq z \neq x$ .Prove that $(x+y+z)(xy+yz+zx-2)\geq 9xyz.$ When does the equality hold?

Since the equation is symmetric and $x,y,z$ are distinct integers WLOG we can assume that $x\geq y+1\geq z+2$ . \begin{align*}

 x+y+z\geq 3(z+1)\\
 xy+yz+xz-2 = y(x+z)+xy-2 \geq (z+1)(2z+z)+z(z+2)-2 \\
 xy+yz+xz-2 \geq 3z(z+2)

\end{align*} Hence $(x+y+z)(xy+yz+xz-2)\geq 9(z)(z+1)(z+2)=$

2017 JBMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4
All JBMO Problems and Solutions

@@ Line 5: / Line 5: @@
 == Solution ==
+Since the equation is symmetric and <math>x,y,z</math> are distinct integers WLOG we can assume that <math>x\geq y+1\geq z+2</math>.
+\begin{align*}
+  x+y+z\geq 3(z+1)\
+  xy+yz+xz-2 = y(x+z)+xy-2 \geq (z+1)(2z+z)+z(z+2)-2 \
+  xy+yz+xz-2 \geq 3z(z+2)
+\end{align*}
+Hence <cmath>(x+y+z)(xy+yz+xz-2)\geq 9(z)(z+1)(z+2)=</cmath>
 == See also ==