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Here excerpt from a book:

Аssume that $f$ satisfies $\nabla f(x) \ge 0$ for all $x$, but is not nondecreasing, i.e., there exist $x,y$ with $x < y$ and $f(y) < f(x)$. By differentiability of $f$ there exists at $t\in[0,1]$ with

$$\frac{d}{dt}f\left(x+t(y−x)\right) =\nabla f\left(x+t\left(y−x\right)\right)^T(y−x)<0.$$

I don't understand why the derivative is less than zero? The function could look like on the image. enter image description here

The book is Convex Optimization by Stephen Boyd and Lieven Vandenberghe, page 109.

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Yola
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Let $g(t)\equiv f(x+t(y-x))$ be defined on $[0,1]$. Then, $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$. So by the Mean Value Theorem, there is some $t^*\in(0,1)$ such that $$ g'(t^*)=\frac{g(1)-g(0)}{1-0}=f(y)-f(x)<0. $$ It remains to expand the leftmost expression above.

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