Third Method of Solving IMO Optimization Contest Problem:
Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.
This third method is provided by Mark, another blog contributor and he approached the problem using the well-known Lagrange multipliers method:
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Second Method of Solving IMO Optimization Contest Problem: Find minimum $xy$
Second Method of Solving IMO Optimization Contest Problem:
Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.
The second method is the solution provided by a well-known mathematics retired professor from the University in the U.K.
Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.
The second method is the solution provided by a well-known mathematics retired professor from the University in the U.K.
IMO Optimization Contest Problem: Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.
IMO Optimization Contest Problem:
Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.
My solution:
From the well-known identity
$(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$
and the given values for $x^2+y^2+z^2=7$ and $xy+xz+yz=4$, we get:
$(x+y+z)^2=7+2(4)$
Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.
My solution:
From the well-known identity
$(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$
and the given values for $x^2+y^2+z^2=7$ and $xy+xz+yz=4$, we get:
$(x+y+z)^2=7+2(4)$
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