Difference between revisions of "2017 JBMO Problems/Problem 2"
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
x+y+z\geq 3(z+1)\\ | x+y+z\geq 3(z+1)\\ | ||
| − | xy+yz+xz-2 = y(x+z)+xy-2 \geq (z+1)(2z+ | + | 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) | xy+yz+xz-2 \geq 3z(z+2) | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
Hence <cmath>(x+y+z)(xy+yz+xz-2)\geq 9(z)(z+1)(z+2)</cmath> | Hence <cmath>(x+y+z)(xy+yz+xz-2)\geq 9(z)(z+1)(z+2)</cmath> | ||
| + | |||
| + | THIS IS A FAKESOLVE: The inequality in the last line proven is a lot weaker than the original inequality. This needs to be fixed. | ||
== See also == | == See also == | ||
Latest revision as of 00:36, 21 June 2024
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 
.
Hence ![\[(x+y+z)(xy+yz+xz-2)\geq 9(z)(z+1)(z+2)\]](http://latex.artofproblemsolving.com/0/6/e/06e19b24b54e55eb1debd2ab894fcfd2e2e078d0.png)
THIS IS A FAKESOLVE: The inequality in the last line proven is a lot weaker than the original inequality. This needs to be fixed.
See also
| 2017 JBMO (Problems • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 | ||
| All JBMO Problems and Solutions | ||
