Is every non-increasing function not increasing?

Question

I want to prove or disprove my hypothesis: Let $f:(a,b)\rightarrow R$ be non-increasing function, then it is not increasing.

Definitions: $$\text{Function is increasing iff $x<y\Rightarrow f(x)<f(y)$}\tag{1}$$

$$\text{Function is non-increasing iff $x<y\Rightarrow f(x)\geq f(y)$}\tag{2}$$ So, my idea was to just write the theorem as it is: $$\{\forall x,y\in(a,b):x<y\Rightarrow f(x)\geq f(y)\}\Rightarrow \{\forall x,y\in(a,b):x<y\Rightarrow f(x)< f(y)\}$$ And make use of truth values of $x<y,f(x)<f(y)$ but I don't get a tautology, I couldn't make use of the quantifiers.

Secondly I would be interested in disproving the converse statement that any not inreasing function is non-increasing. As a counter example i would give $x\mapsto \sin{x}$ on $(0,2\pi)$

Answer 1

Non-increasing:

$$\forall\,x, y: x<y:f(x)\ge f(y).$$

Not increasing:

$$\overline{\forall\,x,y:x<y:f(x)<f(y)}\iff \exists\,x,y:x<y:f(x)\ge f(y).$$

Obviously,

$$\forall\,x, y: x<y:f(x)\ge f(y)\implies \exists\,x,y:x<y:f(x)\ge f(y).$$

Notice the change of the quantifier.

Answer 2

Hint: By trichotomy $a < b \iff a \not \ge b$.