To Show $ x.\nu > 0 $ on convex domain. $\Omega\subset \mathbb{R}^n $ be a convex domain, $\nu$ be the unit outward normal of the smooth boundary, then I have to show that $ x.\nu > 0 $ for all $ x \in \partial \Omega$. 
Here is my try. WLOG we can locally assume some convex function $ f \in C^1(\mathbb{R}^{n-1}) $ such that for some $ x_0 \in \partial \Omega $ we can locally have $ \Omega \cap B(x_0, r) = B(x_0, r) \cap \{ x_n > f(x') \} $ and 
$ \partial \Omega\cap B(x_0, r) = B(x_0, r)\cap \{ x_n = f(x')\} $, where $ x = (x',x_n) \in \mathbb{R}^n $
hence locally we have unit normal given by 
$$ \nu = \frac{(\nabla f,-1)}{\sqrt{1+|\nabla f|^2}}$$
Thus $ x.\nu >0 $ iff $ x'.\nabla f(x') -x_n > 0 $. So for $x = (x', x_n) \in \partial \Omega $ I define $g(t) = f(tx') -tx_n $ which too is convex and $g(0) = f(0), g(1) = 0 $ as $ f(x') = x_n $ in boundary. Hence using convexity I obtain  $ g(0) > g(1) - g'(1) = -g'(1) $. So $ g'(1) + g(0) > 0 $ that is finally I have
$$ x'.\nabla f(x') -x_n + f(0) > 0 $$ 
And here I am stuck, I can't get rid of $f(0)$. Any hints or solutions would be extremely helpful. Thank you.
 A: First note that the hypothesis $0\in\Omega$ is necessary. Indeed, consider the sphere given by $$\{x\in \mathbb{R}^n:\ \|x-(1,1,...,1)\|=1\}$$
From here we will assume that $0\in \Omega$ and a more general condition, to wit, $\Omega$ is a starshaped domain with respect to the origin and $\partial\Omega$ is smooth.
Here is the idea of the proof: consider the continuous function $f:\partial\Omega\to\mathbb{R}$ defined by $$f(x)=x\cdot\nu.$$
I - Show that there is no point in $\partial\Omega$ for which $f(x)=0$. (Hint: try it by contradiction).
II - Show that there is a point $z\in\partial\Omega$ such that $f(z)>0$.
To prove II, consider a hyperplane $P$ in $\mathbb{R}^n$, i.e. a set of the form $\{x+a\in \mathbb{R}^n:\ T(x)=0\}$, where $a\in\mathbb{R}^n$ and $T$ is a linear map. First take $P$ such that $P$ does not intersect $\partial\Omega$. Now you translate $P$ until the moment it touches $\partial\Omega$ for the first time in some point $z$.
$\bf{Claim}$: $f(z)>0$. (Hint: try it by contradiction).
Remark: After adding the hypothesis $0\in\Omega$, we can finish your proof in the following way. Fix $y=(y',y_n)\in\partial\Omega$.
As $0\in\Omega$ and $\Omega$ is convex, we can assume (after a rotation  if necessary) that there is $r>0$ and a convex function $f\in C^1(B(0,r))$ such that $f$ is positive and $$(0,f(0))=y,\ ~~ \ \Omega\cap f(B(0,r))=\{(x',x_n)\in\mathbb{R}:\ f(x)<x_n\},$$
$$\partial\Omega\cap F(B(0,r))=\{(x',x_n)\in\mathbb{R}^n:\ f(x)=x_n\}$$
Now note that $$\nu=\frac{(\nabla f,1)}{\sqrt{1+|\nabla f|^2}}.$$
Therefore, in the point $y$, we must have (remember that $y=(0,f(0))$) $$(0,f(0))\cdot \frac{(\nabla f(0),1)}{\sqrt{1+|\nabla f(0)|^2}}=\frac{f(0)}{\sqrt{1+|\nabla f(0)|^2}}>0.$$
