Lagrange Multiplier Question and my attempt Question is Find the extrema of $xyz$ when $x+y+z=a$ , a>0. Strating with usual Lagrange Multiplier method
i get
$f_x$ = $yz$ +$\lambda$   =0                                
$f_y$ = $xz$ +$\lambda$ =0
$f_z$ = $yz$ +$\lambda$ =0
Now from three equations above of  i multiply first by $x$ and second by $y$ and third by $z$ i get
$f_x$ = $xyz$ +$\lambda(x)$   =0                                
$f_y$ = $xyz$ +$\lambda(y)$ =0
$f_z$ = $xyz$ +$\lambda$(z) =0
Clearly from these equating values of $xyz$ . i get $x=y=z$ .And thus i have solved the question and is consistent with my answer with textbook
BUT,If i manipulate equations in a way as if i equate values of $\lambda$ i get
$yz=xz=xy$
Now i take $yz=xz$ .This implies either $z=0 or x=y$ .
IF i take $z=0$ and put in other two equations i get $xy=0$ which means either x=0 or y=o.Say i take x=0 and now putting in constraint equation i get $y=a$ so i get $(0,a,0)$ ..Not only this but by solving other equations like this i get $(0,a,0) , (a,0,0) , (0,0,a) , ((a-1)/2,(a-1)/2,1)$
but this isnot consistent with textbook .Can anybody help me out from here .Thanks

You have derived the correct result with the Lagrange Multiplier method:
$xy = yz = zx = -\lambda$
Here $\lambda$ is a non-zero parameter. From this it follows that $x, y$ and $z$ can not be zero. The only solution is $x = y = z = \sqrt{-\lambda}$. And therefore $x = y = z = a/3$. 
