If $\inf f = f(a)$, then $\exists b,c$, $f(b) = f(c)$

Question

Let $f$ be a real differentiable function on some $[x,y]$, and suppose that $\inf f = f(a)$ for some $a \in (x,y)$. How do I show that: $\exists b,c$ such that $x \le b < a < c \le y$ and $f(b) = f(c)$? It is somehow intuitively true, however I'd like to see a formal proof.

Edit: Suppose that if the infimum is not at a unique point, then we may choose $a$ to be any one of these points.

Answer 1

Disproved: $$f(x)=\begin{cases}x^2&x\ge0\\0&x\le0\end{cases}$$ Let $[x,y]$ be $[-1,1]$ and $\inf f=0=f(0)$, taking $a$ as $0$. Now $-1\le b<0<c\le1$ and $f(b)=f(c)\implies c^2=0\implies c=0$ which is a contradiction. $\hskip2in$

Answer 2

If your allowed to choose $a$ then your statement is true. To see this let us assume that an initial $a_0$ with the given property is chosen. Then if $f$ is constant in a small neighborhood of $a$ then we are done, otherwise WLOG let us assume that $f(c)>f(a)$ for $c>a_0$ and $c$ is in some small neighborhood of $a_0$. If we assume that for all $b$, where $x<b<a_0$ $f(b)=f(a_0)$, then we can choose $a=\frac{a_0-x}{2}$ and we can choose $x<b<a<c<a_0$ such that the statement is true. If however there does exist $b$ such that $f(b)>f(a_0)$ then we know that either $f(b)\leq f(c)$ or $f(c)\geq f(b)$. In whichever case it is we choose the smaller number. Say WLOG it is $f(b)$. Then by intermediate value theorem there exists $a_0<c'<c$ such that $f(c')=f(b)$. This completes the proof.