Path connectedness of a particular algebraic set Let $n\geq 3$, let $a$ and $b$ be two positive numbers, and
$$
\Omega = \bigg\lbrace (x_1,x_2, \ldots ,x_n) \in ]0,+\infty[^n \ \bigg| \  x_1+x_2+ \ldots +x_n=a, \ x_1x_2 \ldots x_n=b \bigg\rbrace
$$
It is easy to see that $\Omega$ is nonempty iff $\frac{a}{n} \geq b^{\frac{1}{n}}$.
It seems clear enough that $\Omega$ is path connected, but I could find no simple proof of that.
 A: Assume $\frac{a}{n} \geq b^{1/n}$ so that $\Omega \neq \emptyset$.
Let $H$ be the hyperplane of $\mathbb{R}^n$ defined by the equation $x_1+x_2+\cdots +x_n = a$, and let $B = \{ (x_i) \in (0, \infty)^n \;|\; x_1x_2 \cdots x_n \geq b\}$.  It should be clear that $\Omega = H \cap \partial B = \partial(H \cap B)$.  Let's first consider the space $U = H \cap B$.  You can show that $B$ is convex of dimension $n$, and since $H$ is a hyperplane, we have that $U$ is convex of dimension $\leq n-1$, hence path-connected.  Furthermore, $U$ is bounded, since $H \cap B \subseteq (0,a)^n$ 
Now if $dim(U) < n-1$, then $U$ is a single point because $\partial B$ contains no line segment (see below for more details on this).  In that case, $\Omega$ is definitely connected.  On the other hand, when $dim(U) = n-1$, convexity and boundedness imply that $\partial U$ is homeomorphic to $S^{n-2}$.  Since you have $n \geq 3$, we find that $\Omega = \partial U$ is connected.
ADDENDA:
Lemma:  In $\mathbb{R}^n$, If a hyperplane $H$ intersects a convex set $B$ of dimension $n$ in such a way that there is an interior point $x$ of $B$ in the intersection, then $dim(H \cap B) = n - 1$.
Proof:  If $x$ is interior to $B$, then there is an $n$-ball $V_x \subset B$ containing $x$.  Suppose $V_x = \{ y \in \mathbb{R}^n \;|\; dist(x,y) \leq r\}$.  Now $V_x \cap H = \{ y \in H \;|\; dist(x,y) \leq r\}$.  Thus, $V_x \cap H$ is an $(n-1)$-dimensional ball in $H \cap B$.
Cor:  If $dim(H \cap B) < n-1$, then $H \cap B \subseteq H \cap \partial B$.
Lemma: If $B \subseteq \mathbb{R}^n$ is convex and $H$ is a hyperplane such that $x \in H \cap B \subseteq H \cap \partial B$, and $\partial B$ contains no line segment, then $H \cap \partial B = \{ x\}$.
Proof: If $y \neq x$ is another point of $H \cap \partial B$, then we have two points $x, y \in H \cap B$, a convex set.  Thus the line segment $\ell$ joining $x$ and $y$ is in $H \cap B \subseteq H \cap \partial B$.  In particular, $\ell \in \partial B$, contradicting that $\partial B$ contains no segment.
