Suppose we have the following linear programming problem in standard form:
$$ \min c^Tx $$ $$ \text{s.t.}\quad Ax=b, x\ge 0 $$
Make the above problem unconstrained via the Lagrangian method, with multiplier $p$
$$\min c^Tx + p^T(b-Ax) $$ $$ \text{s.t.}\quad x\ge 0$$
My question:
Why does the relaxed (i.e. Lagrangian) problem's value function is weakly less than the value function of the original problem (see below)? Can someone explain this intuitively or geometrically?
$$h(p)=\min\,[c^Tx+p^T(b-Ax)]\,\leq\, c^Tx^*+p^T(b-Ax^*)=c^Tx^*$$
Let
$$ L(x, \lambda) := c^T x + \lambda(b-Ax) $$
then you can define a function $g$ with
$$ g(\lambda):= \min_{x\ge 0} L(x,\lambda) $$
and, by definition of this function, $g(\lambda) \le L(x,\lambda)$ for any $x\ge 0$ and any $\lambda$ (this follows from having the "min" in the definition of $g$)
Now the optimum for the initial problem (if it exists) verifies $x^*\ge 0$ so that it must verify the inequality:
$$ g(\lambda) \le L(x^*, \lambda) $$
Now, since it must also verifies $Ax^*=b$, you have $L(x^*,\lambda)=c^Tx^*$.
Making things explicit, you recover the inequality:
$$ \min_{x\ge0}\,\, [c^T+\lambda(b-Ax)] \le c^T x^*.$$