Difference between revisions of "2014 UMO Problems/Problem 5"

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== Solution ==
 
== Solution ==
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By AM-GM
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<cmath>x^2+y^2+z^2 + 2yz\ge x^2 + 4yz\ge 4x\sqrt{yz}</cmath>
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Hence, the second equation implies that <math>y=z</math> and <math>x^2=4yz\implies x=2y=2z</math>.
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Now we plug it into the first equation to get <math>(x,y,z) = \left(\sqrt[3]{4}, \frac{\sqrt[3]4}{2}, \frac{\sqrt[3]4}{2}\right)</math>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 15:48, 12 February 2020

Problem

Find all positive real numbers $x, y$, and $z$ that satisfy both of the following equations. \begin{align*} xyz & = 1\\ x^2 + y^2 + z^2 & = 4x\sqrt{yz}- 2yz \end{align*}

Solution

By AM-GM \[x^2+y^2+z^2 + 2yz\ge x^2 + 4yz\ge 4x\sqrt{yz}\] Hence, the second equation implies that $y=z$ and $x^2=4yz\implies x=2y=2z$.

Now we plug it into the first equation to get $(x,y,z) = \left(\sqrt[3]{4}, \frac{\sqrt[3]4}{2}, \frac{\sqrt[3]4}{2}\right)$

See Also

2014 UMO (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6
All UMO Problems and Solutions
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