Belarusian Mathematical Olympiad

Answer: $x=1$.

First, we prove that if $a$, $b$, $c$ are the sides of a triangle, then the inequality $$x+c\le (x+a)(x+b)(\ast )$$ 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 $\frac{1-x}{2}$.