Third Method of Solving IMO Optimization Contest Problem: Find the minimum value of $xy$,

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:

We are given the objective function:

[MATH]f(x,y,z)=xy[/MATH]

subject to the constraints:

[MATH]g(x,y,z)=x^2+y^2+z^2-7=0[/MATH]

[MATH]h(x,y,z)=xy+xz+yz-4=0[/MATH]

Using Lagrange Multipliers, we obtain:

[MATH]y=\lambda(2x)+\mu(y+z)[/MATH]

[MATH]x=\lambda(2y)+\mu(x+z)[/MATH]

[MATH]0=\lambda(2z)+\mu(x+y)[/MATH]

Adding, we then get:

[MATH]x+y=2(x+y+z)(\lambda+\mu)[/MATH]

The third equation gives us:

[MATH]\mu=-\lambda\frac{2z}{x+y}[/MATH]

Hence:

[MATH]\lambda=\frac{(x+y)^2}{2(x+y+z)(x+y-2z)}[/MATH]

[MATH]\mu=-\frac{z(x+y)}{(x+y+z)(x+y-2z)}[/MATH]

And so, using the first equation, we find:

[MATH]y=\frac{x(x+y)^2}{(x+y+z)(x+y-2z)}-\frac{z(x+y)(y+z)}{(x+y+z)(x+y-2z)}[/MATH]

[MATH]y(x+y+z)(x+y-2z)=x(x+y)^2-z(x+y)(y+z)[/MATH]

This reduces to:

[MATH]z^2=(x+y)^2[/MATH]

Using the second equation, we obtain:

[MATH]x(x+y+z)(x+y-2z)=y(x+y)^2-z(x+y)(x+z)[/MATH]

This results in complex values for $z$.

Thus, the first constraint becomes:

[MATH]2x^2+2xy+2y^2-7=0[/MATH]

And from the second, we find:

[MATH]2x^2+6xy+2y^2-8=0[/MATH]

Subtracting the first of these last two results from the second, we obtain:

[MATH]4xy=1\implies xy=\frac{1}{4}[/MATH]