Using technique of Lagrange Multiplier

Question

How can we use the technique of Lagrange multipliers to find a new vector of parameters $w$ which solves the optimization problem:

minimize J(w) = $\frac{1}{2} || w -u ||^2$

such that:
$w^T (x − y) ≥ 1$

Here $u$ is the current vector of parameters and $x$ and $y$ are two training examples such that $u^T(x-y) < 1$.

My attempt:

I rewrote the constraints as
$1 -wx -wy \leq 0 $
$ ux -uy -1 < 0$

Then the Lagrangian is,
$L = \frac{1}{2} (w-u)^2 + a(1-wx-wy) + b(ux-uy-1)$
where, $a,b \geq 0 $ are Lagrange multipliers.

But, when I set gradient to zero:
$0 = \frac{dL}{dw} = \frac{dL}{du} = \frac{dL}{dx} = \frac{dL}{dy} = \frac{dL}{da} = \frac{dL}{db} $

I got $w^* = 0 $, but I have to solve for w and I assume zero is not the correct answer here.

What am I doing wrong?

Update
Following the idea of jbowman, I got this,
$z = x-y$
minimize J(w) = $\frac{1}{2} || w -u ||^2$

with constraint $1 - wz \leq 0$
Primal Lagrangian $L_P = \frac{1}{2} (w-u)^2 + a(1-wz) $
$0 = \frac{dL}{dw} = \frac{dL}{du} = \frac{dL}{dz} = \frac{dL}{da} $

Another attempt,
I did not include dL/du and take only
$0 = \frac{dL}{dw} = \frac{dL}{dz} = \frac{dL}{da} $

Then I got
Dual Lagrangian,
$L_D = infimum_a L_P(w,u,z,a) = \frac{1}{2} a^2 z^2 +a $

Again, I got w = u.

This is problem of Large Margin Perceptron, but I got answer w equal to u. Whatever I initialize the values to u, the final weight vector w will be u.

Certainly, something is wrong. I am wondering what I did wrong?

Answer 1

Following the suggestion of jbowman, I derived the gradient w.r.t. only w and a and got the quadratic solution for w. Optimization problem:

minimize J(w) = $\frac{1}{2} || w -u ||^2$

such that:
$w^T (x − y) ≥ 1$

Constraint:
$ 1 - wz \le 0 $, where z = x-y.

Primal Lagrangian:
$L_P = \frac{1}{2} (w-u)^2 + \alpha(1-wz) $ where $\alpha \ge 0 $ is the Lagrange multiplier.

Set gradient w.r.t. w and a to zero:
$0 = \frac{dL}{dw} ==> w = u + \alpha z $
$0 = \frac{dL}{d\alpha} ==> z = 1/w $

Solving these two equations, I got

$w^* =\frac{u \ \pm \sqrt(u^2 + 4\alpha) }{2} $

This is the answer I came with, but not sure its right or not.

Supplementary materials: