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

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    $\begingroup$ Show your efforts in an edit. $\endgroup$
    – J.G.
    Aug 27, 2019 at 18:38

3 Answers 3

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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.

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  • $\begingroup$ Sir ..can u please help me to find x=a $\endgroup$
    – Cent22
    Aug 27, 2019 at 19:06
  • $\begingroup$ And i got lambda equal to 1 $\endgroup$
    – Cent22
    Aug 27, 2019 at 19:07
  • $\begingroup$ @Cent22 Sorry, you're right about $\lambda$. But my edit explains why $x=y=z$. $\endgroup$
    – J.G.
    Aug 27, 2019 at 19:09
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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$

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    $\begingroup$ 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. $\endgroup$
    – J.G.
    Aug 27, 2019 at 18:42
  • $\begingroup$ @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. $\endgroup$ Aug 27, 2019 at 18:49
  • $\begingroup$ @J.G. Your comment is very helpful. I edited my solution accordinly. $\endgroup$ Aug 27, 2019 at 19:13
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\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}

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  • $\begingroup$ Thanks for ur kind help...but how (a)+(b) gives 4x=x+3z $\endgroup$
    – Cent22
    Aug 27, 2019 at 19:15
  • $\begingroup$ \begin{align} 2x &= y+z\\ 4x &= 2y + 2z\\ 4x &= x+z + 2z\\ 4x &= x + 3z \end{align} $\endgroup$
    – John Lou
    Aug 27, 2019 at 21:35
  • $\begingroup$ Thank you...i understand now ....... $\endgroup$
    – Cent22
    Aug 28, 2019 at 2:21

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