$x = 1$.

First, we prove that if $a$, $b$, $c$ are the sides of a triangle, then the inequality
$$x + c \le (x + a)(x + b) \quad (*)$$
holds for $x = 1$. Indeed, we can rewrite $(*)$ as $x + c \le x^2 + (a + b)x + ab$. It is easy to see that this inequality holds for $x = 1$ since $x^2 = x = 1$ and, by the triangle inequality, $(a + b)x = a + b > c$.

Now we show that for any $x < 1$ there exists a triangle such that $(*)$ does not hold. Indeed, if $x < 0$, then it suffices to consider the triangle with the sides $a = 1 - x$, $b = 1 - x$, and $c = 2 - x$. If $0 \le x < 1$, then it suffices to consider the regular triangle with the side $\frac{1 - x}{2}$.

Answer: $x = 1$.

First, we prove that if $a$, $b$, $c$ are the sides of a triangle, then the inequality
$$x + c \le (x + a)(x + b) \quad (*)$$
holds for $x = 1$. Indeed, we can rewrite (*) as $x + c \le x^2 + (a + b)x + ab$. It is easy to see that this inequality holds for $x = 1$ since $x^2 = x = 1$ and, by the triangle inequality, $(a + b)x = a + b > c$.

Now we show that for any $x < 1$ there exists a triangle such that (*) does not hold. Indeed, if $x < 0$, then it suffices to consider the triangle with the sides $a = 1 - x$, $b = 1 - x$, and $c = 2 - x$. If $0 \le x < 1$, then it suffices to consider the regular triangle with the side $\frac{1 - x}{2}$.