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From Boyd and Vanderbei's Convex Optimization, a differentiable function $f$ is convex iff $dom f$ is convex and for all $x,yin dom f$,

$$f(y)ge f(x)+nabla f(x)^T(y-x).$$

I am having trouble following their proof that this latter condition implies convexity, assuming $f$ is a single variable function. I will outline their argument and explain where I fall off:

They start out by picking $thetain[0,1]$, and $x,yin dom f$. If $f$ is a single-variable function, the assumed equality tells us that $theta f(x)ge theta f(z)+theta f'(z)(x-z)$ and $(1-theta)f(y)ge (1-theta) f(z)+(1-theta)f'(z)(y-z)$. We are supposed to add these inequalities to show $f$ is convex, i.e. $theta f(x)+(1-theta) f(y)ge f(z)$. So it must be the case that $theta f'(z)(x-z)+(1-theta)f'(z)(y-z)ge 0$. I'm not able to show this.

Any help is greatly appreciated. Thanks in advance.

The estimates are applied to $x, y$, and $z= theta x + (1-theta) y$.
Adding the inequalities gives
$$
theta f(x) + (1-theta) f(y)ge f(z) + bigl(underbrace{theta (x-z) + (1-theta) (y-z)}_{=0}bigr) f'(z) = f(z) , .
$$