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

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== Solution ==
 
== Solution ==
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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>.
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<cmath>\begin{align*}
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  x+y+z\geq 3(z+1)\\
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  xy+yz+xz-2 = y(x+z)+xy-2 \geq (z+1)(2z+2)+z(z+2)-2 \\
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  xy+yz+xz-2 \geq 3z(z+2)
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\end{align*}</cmath>
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Hence <cmath>(x+y+z)(xy+yz+xz-2)\geq 9(z)(z+1)(z+2)</cmath>
  
 
== See also ==
 
== See also ==

Latest revision as of 10:05, 29 July 2022

Problem

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?

Solution

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+2)+z(z+2)-2 \\   xy+yz+xz-2 \geq 3z(z+2) \end{align*} Hence \[(x+y+z)(xy+yz+xz-2)\geq 9(z)(z+1)(z+2)\]

See also

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