# How to find minimum value by Lagrange multiplier method? [closed]

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

• Show your efforts in an edit.
– J.G.
Aug 27, 2019 at 18:38

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.

• And i got lambda equal to 1 Aug 27, 2019 at 19:07
• @Cent22 Sorry, you're right about $\lambda$. But my edit explains why $x=y=z$.
– J.G.
Aug 27, 2019 at 19:09

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 $$(3a)^2-6a^2=3a^2$$

• So you're saying a sum of three real squares can go down as far as $-6a^2$? Not if $a>0$ it can't. Your problem is in assuming $x+y+z=0,\,xy+yz+zx=3a^2$ has simultaneous solutions in the reals; but you can show this reduces to $y^2+yz+z^2=-3a^2$, again impossible.
– J.G.
Aug 27, 2019 at 18:42
• @J.G. So what this achieves is to show that one can attack the problem by minimising the value of $|x+y+z|$ subject to the constraint. Which could be easier. Aug 27, 2019 at 18:49
• @J.G. Your comment is very helpful. I edited my solution accordinly. Aug 27, 2019 at 19:13

\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}

• Thanks for ur kind help...but how (a)+(b) gives 4x=x+3z Aug 27, 2019 at 19:15
• \begin{align} 2x &= y+z\\ 4x &= 2y + 2z\\ 4x &= x+z + 2z\\ 4x &= x + 3z \end{align} Aug 27, 2019 at 21:35
• Thank you...i understand now ....... Aug 28, 2019 at 2:21