Lagrange's method question

Question

Find the extreme values of the function $f(x,y,z)=x^2+y^2+z^2$ subject to the condition $xy+yz+zx=3a^2$.

I tried to solved it by Lagrange method and got $3$ equations.

\begin{align*} &2x+k(y+z) = 0 \\ &2y + k(x+z) = 0 \\ &2z + k(x+y) = 0 \end{align*}

Now to get the value of $k$, I multiplied first equation by $x$, second equation by $y$ and third equation by $z$. And added all the $3$ equations. I got the value of $k$. $$k= -(x^2+y^2+z^2)/3a^2$$

Now how do I use this value of $k$ to get the values of $x, y, z$. The answer is $(a, a, a)$ and $(-a, - a, - a)$. Help please !

Answer 1

you have these 4 equations

$2x+k(y+z)=0$

$2y+k(x+z)=0$

$2z+k(x+y)=0$

$xy+yz+zx=3a^2$

find $k$ from the first:

$k=\frac{-2x}{y+z}$

replace it in the second and third

$2y(y+z)=2x(x+z)$

$2z(y+z)=2x(x+y)$

$y(x+z)+xz=3a^2$

rewriting the first and second

$(y-x)(x+y+z)=0$

$(z-x)(x+y+z)=0$

$xy+xz+yz=3a^2$

from the two first we have $x=y=z$, replacing in the third

$x=y=z=a$

Answer 2

HINT: Directly by symmetry $ x=y=z=a. $