$f$ convex $\iff f(y)\geq f'(x)(y-x)+f(x).$ I want to prove that $f$ is convex $\iff f(y)\geq f'(x)(y-x)+f(x)$.
The implication is fine, but I have difficulties with the converse implication. I tried to prove that $f'$ is increasing, but still, it doesn't work : Let $y>x$, then $$\frac{f(y)-f(x)}{y-x}\geq f'(x),$$
but taking $x\to y$ we get $f'(y)\geq \lim_{x\to y}f'(x).$ So if $f$ is $\mathcal C^1$ we don't get better than $f'(y)\geq f'(y)$... any idea ? 

Choose any $x\ne y$, and $0\le\theta\le1$, and let $z=\theta x+(1-\theta)y$. Applying the inequality  twice yields
$$
 f(x)\ge f(z)+f'(z)(x-z)
 \quad\text{and}\quad
 f(y)\ge f(z)+f'(z)(y-z).
$$
Multiplying the first inequality by $\theta$, the second by $1-\theta$, and adding them yields
$$
 f(z)\le\theta f(x)+(1-\theta)f(y),
$$
which proves that $f$ is convex.
