Convexity definition (and intuition) for differentiable function The classic definition of a convex function i.e.
$$
f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda) f(y) \quad 0 \leq \lambda \leq 1 
$$
is crystal clear for me.
However, I'm struggling to find an intuitive interpretation when the function is differentiable:
$$
f(y) \geq f(x) + \langle \nabla f(x), (y - x)\rangle
$$
I know that the RHS is essentially the first order Taylor approximation, but that doesn't help me.
Could someone shed some light on what is intuitively going on here? and why Taylor appears here?
 A: For intuition, let's think in $\mathbb{R}$ first.  Then this is saying that the graph of $f$ lies above its tangent lines. Fix any $x$, and consider the right-hand side as a function of $y$: it's precisely then the graph of the tangent line to $f$ at $x$, and the inequality states that $f(y)$ lies above this line.
In higher dimensions, the exact same intuition holds.  Fix $x$, then the right hand side, as a function of the vector $y$, is just the tangent hyperplane to the graph of $f$ at $x$, and the inequality states that $f(y)$ lies above this hyperplane everywhere, for every $x$.
Edit: To make sense of the inner product, let's first start in $\mathbb{R}$ again.  Then for fixed $x$, the right-hand side, as a function of $y$ is affine.  To see that this is the graph of the tangent line, note the graph of $f'(x)\cdot y$ is just a line with slope $f'(x)$ through the origin.  Then $f'(x) \cdot y + f(x)$ is the same line, but this time with $y$-intercept given by $f(x)$.  Call $g(y) = f(x) + f'(x)\cdot y$.  Then your right-hand side is just $g(y-x)$, which takes the graph of $g$ and shifts it right by $x$.  So the graph of $g$ is just the affine function that takes the value $f(x)$ at $x$, instead of at $0$.
In multiple dimensions, every real valued affine function is of the form $h(x) = \langle a, x\rangle + b$, for vector $a$ and scalar $b$.

Proof: Suppose first $h: \mathbb{R}^N \to \mathbb{R}$ is linear. Now, for all $x \in \mathbb{R}^N$, $x = \sum_{i=1}^N x_i e^i$, where $e^i$ is the $i$-th standard basis vector.  Since $h$ is linear:
$$
h(x) = h\bigg(\sum_{i=1}^N x_i e^i\bigg) = \sum_{i=1}^N h\big(x_i e^i\big) = \sum_{i=1}^N x_i h(e^i).
$$
Thus defining $a = \big[h(e^1), \ldots, h(e^N)\big]$ we see $h(x) = \langle a , x \rangle$.  Suppose now instead of linear $h$ is affine.
Define $b =  h(0)$, and get $a$ from the linear function $h(x) - b$.

So the dot product is just giving you the exact same structure you saw in the single dimensional case, it's defining you an affine function, here the tangent hyperplane to the graph of $f$ at $x$.
