How to find minimum value by Lagrange multiplier method?

Question

I had solved some problems on Lagrange multiplier method but I'm stuck in this question.

Findimum value of $$x^2+y^2+z^2$$ when $$yz+zx+xy=3a^2$$

Answer 1

Note that $$x^2+y^2+z^2= (x+y+z)^2-2(xy+xz+yz)=$$

$$(x+y+z)^2-2(3a^2)=(x+y+z)^2-6a^2$$

Apply Lagrange Multiopliers and yu get $x=y=z=a$

The minimum value is $ (3a)^2-6a^2=3a^2$

Answer 2

With Lagrange multiplier $\lambda$, the Lagrangian is $L=x^2+y^2+z^2+\lambda(3a^2-yz-zx-xy)$. Then $0=\partial_xL=2x-\lambda(y+z)\implies 2x=\lambda(y+z)$. We get two other equations similar to this one; adding all gives $(\lambda-1)(x+y+z)=0$, so the minimum is $3a^2$ with $\lambda=1,\,x=y=z=a$. (Note in particular that if without loss of generality $x\ge y\ge z$ we get $y+z\ge z+x\ge x+y$, so to avoid contradiction $x=y=z$.) An alternative approach is to note we're minimising the squared length of a vector, given its dot product with a rearrangement of its entries; then we can use Cauchy-Schwarz to show the entries need to be equal.

Answer 3

\begin{align} \nabla f(x,y,z) &= \lambda \nabla g(x,y,z)\\ g(x,y,z) &= k\\ 2x &=\lambda(y+z) &\text{(a})\\ 2y &=\lambda(x+z)&\text{(b})\\ 2z &=\lambda(x+y)&\text{(c})\\ xy + xz + yz &= 3a^2\\ 2x + 2y + 2z &= \lambda(2x + 2y + 2z) &\text{(a)+(b)+(c)}\\ \text{Therefore } \lambda &=1\\ 4x &=x+3z &\text{(a)+(b)}\\ \text{With a little more algebra, }x&=z=y = a\\ \text{Therefore the minimum of } x^2 + y^2 + z^2 &= 3a^2\\ \end{align}