Prove that if $a, b, c$ are distinct points of $\mathbb R^d$ such that $∠abc < π/2$, then there is a $x$ on the open segment $(b, c)$ with $|a − x| < |a − b|$.
I tried to take these points as three vertices of an acute-angled triangle such that $∠abc < \pi/2$, and tried to take the property of sum of two sides is greater than the third side i.e. $ab +bc > ac$ then I draw a line from the point $a$ which intersect $bc$ at $x$. Then again I took the sum of two sides property such that $ab + ax > bx$ and after trying all the possibilities I am not unable to prove the above.
Can someone please help me? Thanks in advance.
Consider the triangle $abx$. You are asking that $a$ is closer to $x$ than it is to $b$, or in other words that the line segment $ax$ is shorter than the line segment $ab$. This would be true if and only if the measure of $\angle x$ is greater than the measure of $\angle b$.
There is certainly a point $x$ on the line segment $bc$ that satisfies that. As noted in comments, the projection would work. Alternatively, you could just choose an $x$ so close to $b$ that $m\angle a$ is insignificant, which would make $\angle x$ the largest angle in the triangle (since we are given that $\angle b$ is acute).