**NEW ANSWER**

Now that the problem has been clarified somewhat, under the particular constraint mentioned here, (but not others of a similar nature), the two problems do have the same solution. Let $K,n,\alpha$ be positive with $\alpha \gt \frac{K}{n}$ and consider these optimization problems over non-negative $n$-tuples.

**P1 Maximize $\prod _{i=1}^n x_i$ subject to $\sum _{i=1}^n x_i =K$ AND $x_1 \gt \alpha.$**

**P3: Minimize $\sum _{i=1}^n x_i^2$ subject to $\sum _{i=1}^n x_i =K$ AND $x_1 \gt \alpha$.**

The common extreme is at $x_1=\alpha$ and $x_2=x_3=\cdots=x_n=\frac{K-\alpha}{n-1}.$ To see this, consider any other $n$-tuple. If it has $x_1 \gt \alpha$ then the minimum entry is $x_n \lt \frac{K-\alpha}{n-1}.$ Replace $x_1$ by $x_1-\epsilon$ and $x_n$ by $x_n+\epsilon$ where $\epsilon=\min(x_1-\alpha,\frac{K-\alpha}{n-1}-x_n)$. Observe that this increases what we want to maximize in **P1** and decreases what we want to minimize in **P2**. In the case that $x_1=\alpha$ but $x_2 \gt \frac{K-\alpha}{n-1} \gt x_n$ we replace $x_2$ by $x_2-\epsilon$ and $x_n$ by $x_n+\epsilon$ where now $\epsilon=\min(x_2-\frac{K-\alpha}{n-1},\frac{K-\alpha}{n-1}-x_n).$ After at most $n$ (strict) improvements by $\epsilon$-moves of this sort we arrive at the claimed $n$-tuple.

Consider non-negative triples with $x_1+x_2+x_3=4.46$ and the constraint that at least one of $x_1 \ge 2$ or $x_3 \le 1$ must be true. By an $\epsilon$ argument as above, a maximum for **P1** is at either $\mathbf{x}=(2,1.23,1.23)$ or $\mathbf{y}=(1.73,1.73,1)$ and the same is true of a minimum for **P3**. It turns out that these extremes do not occur at the same place: the maximum for **P1** is at $\mathbf{x}$ since $3.0258 \gt 2.9929$ but the minimum for **P3** is at $\mathbf{y}$ since $7.0258 \gt 6.9858$

Notes:

As far as the result, we could also have used this constraint:
*at least one of* $x_1,x_2,x_3$ *is an integer.*

For this example to work with $K$ slightly larger than $4$ requires $4.414 \lt 3+\sqrt{2} \lt K \lt 4.5.$ In my experience (see the example in my old answer below) things are this delicate and using a larger $K$ makes them even more delicate. Another example with $n=3$ is $K=1.2$ with contenders $(0,0.6,0.6)$ and $(0.1,0.1,1).$

General Comments

Dropping the new constraints, The first problem can be equivalently rewritten as

**P1 Maximize $\sum _{i=1}^n \ln(x_i)$ subject to $\sum _{i=1}^n x_i =K$.**

Then it has the form: Maximize $\sum _{i=1}^n f(x_i)$ subject to $\sum _{i=1}^n x_i =K$ where $f$ is a *concave* function in the sense that $f(u+\epsilon)+f(v-\epsilon) \gt f(u)+f(v)$ provided $u \lt u+\epsilon \lt v-\epsilon \lt v.$ Other examples include $f(x)=\arctan(x)$ and $f(x)=x^p$ with $0 \lt p \lt 1.$ Note than $\ln(x)=\lim \frac{x^p-1}{p}$ as $p$ approaches $0$ from above.

And **P2** has the form: Minimize $\sum _{i=1}^n g(x_i)$ subject to $\sum _{i=1}^n x_i =K$ where $g$ is a *convex* function in the sense that $g(u+\epsilon)+g(v-\epsilon) \lt g(u)+g(v)$ provided $u \lt u+\epsilon \lt v-\epsilon \lt v.$ Here we have essentially the $\ell_2$ norm and $g(x)=|x|^p$ for $p>1$ gives the $\ell_p$ norm. Another example is $g(x)=x\ln(x)$

Call a replacement of $u \lt v$ by $u+\epsilon \lt v-\epsilon$ an $\epsilon$-*move* and call an $n$-tuple $\epsilon$-*optimal* if no move of this type is possible (subject to any further constraints).
Then any optimization problem as above will attain its extreme value at an $\epsilon$-optiml tuple. So all achieve their extreme value at $x_1=\cdots=x_n=\frac{K}{n}$ if that is possible. When there are several $n$-tuples which are optimal in that no further $\epsilon$ moves are possible, then it may be that some optimize one function and others optimize another.

**OLD ANSWER**
I will alter the question and then give an answer of NOT ALWAYS.

I will take the question to be this: Given an $n$-tuple $\mathbf{x}=x_1,x_2,\cdots,x_n$ with $K=\sum _{i=1}^n x_i$, let $\alpha(\mathbf{x})=\Pi _{i=1}^n x_i$ and $\beta(\mathbf{x})=\sum _{i=1}^n (x_i-K/n)^2.$ Suppose we have a set of positive $n$-tuples all with the same sum $K$. We can define a preorder on them by $\mathbf{x} \ge \mathbf{y}$ when $\alpha(\mathbf{x}) \ge \alpha(\mathbf{y})$ OR by $\mathbf{x} \ge \mathbf{y}$ when $\beta(\mathbf{x}) \le \beta(\mathbf{y})$.

Question: does $\alpha(\mathbf{x}) \le \alpha(\mathbf{y})$ always imply $\beta(\mathbf{x}) \ge \beta(\mathbf{y})?$

The question is really about the maximal elements in these two orders but we can always make rules to eliminate the ties.

I wish I had a more natural seeming example but: Consider these triples with sum $30$: $\mathbf{x}=[8.1,10.95,10.95]$ and $\mathbf{y}=[9,9,12]$ then $\alpha(\mathbf{x})=971.21025<972=\alpha(\mathbf{y})$ but also $\beta(\mathbf{x})=5.415<6=\beta(\mathbf{y}).$

Whatother constraints? – Yemon Choi May 8 '11 at 3:53