If $\Omega\subset\mathbb{R}^n$ is convex and $f$ is differentiable, then $f(x)-f(y)\geq f'(y)(x-y),\;\forall \;x,y\in \Omega$ Let $\Omega$ be a convex set in $\Bbb{R}^n$. We say that that $f:\Omega\to \Bbb{R}$ is convex if $$f(tx+(1-t)y)\leq tf(x)+(1-t)f(y),\;\forall\;0\leq t\leq 1,\;\&\;\forall\;x,y\in \Omega.$$ I want to show that if $f$ is convex and  differentiable on $\Omega,$ then $$f(x)-f(y)\geq f'(y)(x-y),\;\forall \;x,y\in \Omega.$$
I'm thinking of using Partial derivatives but don't know how to go about it. Please, can anyone help out?

Clearly You have $$\frac{f(tx+(1-t)y)-f(y)}{t}\leq f(x)-f(y)$$
and for $t\rightarrow 0$ the LHS numerator and denominator tend to zero so that De L´Hospital applies and the partial derivative $\partial/\partial t$ of the numerator is $$f^´(tx+(1-t)y)(x-y)$$ and that of the denominator $1$ and now take $t\rightarrow 0$
