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?
 A: If $f$ is convex and differentiable on the convex set $\Omega$ then it is true that
$$f(y) + \langle \nabla f(y), x-y \rangle \le f(x),$$
for $x,y \in \Omega$.
[Some intuition: The left-hand side is a linear approximation of $f$ near $y$, so geometrically this inequality is saying that the linear approximation lies entirely below the function $f$.]
Proof sketch:
Fix $x$ and $y$ and consider the one-dimensional function $g: \mathbb{R} \to \mathbb{R}$ defined by $g(t) := f(tx + (1-t) y)$. Note that it inherits the convexity property from $f$. Since $g'(t) = \langle\nabla f(tx + (1-t) y), x-y\rangle$, it suffices to show $g(0) + g'(0) \le g(1)$.
Indeed,
$$g'(0) = \lim_{h \downarrow 0} \frac{g(h) - g(0)}{h} \le \lim_{h \downarrow 0}\frac{h g(1) + (1-h) g(0) - g(0)}{h} = g(1) - g(0).$$
A: Let $x,y \in \Omega$, $t \in [0,1]$ and 
$$ g(t)=f(tx+(1-t)y)-tf(x)-(1-t)f(y)$$ 
the convexity of $f$ imply that $g(t) \leq 0$ for every  $t \in [0,1]$ and $g(0)=0 $  so 
$$\frac{g(t)-g(0)}{t} \leq 0  \quad \forall t \in ]0,1] $$
 now taking the limit when $t \rightarrow 0^+$ we have 
$$ g'(t)= f'(y)(x-y)-f(x)+f(y) \leq 0 $$
So, here you are!
A: 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 $t\rightarrow 0$
