2017 JBMO Problems/Problem 2: Difference between revisions
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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 ![\[(x+y+z)(xy+yz+zx-2)\geq 9xyz.\]](http://latex.artofproblemsolving.com/5/6/3/56366cbef963b44b24f2b97c3db0fc34d36dffab.png)
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 ![\[(x+y+z)(xy+yz+xz-2)\geq 9(z)(z+1)(z+2)\]](http://latex.artofproblemsolving.com/0/6/e/06e19b24b54e55eb1debd2ab894fcfd2e2e078d0.png)
See also
| 2017 JBMO (Problems • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 | ||
| All JBMO Problems and Solutions | ||
