Difference between revisions of "2017 JBMO Problems/Problem 2"
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xy+yz+xz-2 \geq 3z(z+2) | xy+yz+xz-2 \geq 3z(z+2) | ||
\end{align*} | \end{align*} | ||
− | Hence <cmath>(x+y+z)(xy+yz+xz-2)\geq 9(z)(z+1)(z+2) | + | Hence <cmath>(x+y+z)(xy+yz+xz-2)\geq 9(z)(z+1)(z+2)</cmath> |
== See also == | == See also == |
Revision as of 08:27, 24 June 2019
Problem
Let be positive integers such that .Prove that When does the equality hold?
Solution
Since the equation is symmetric and are distinct integers WLOG we can assume that . \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
See also
2017 JBMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 | ||
All JBMO Problems and Solutions |