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We know that a function is convex if we have $$\lambda f(x_1) + (1-\lambda)f(x_2) \ge f(\lambda x_1 + (1-\lambda)x_2)$$ where $0\le\lambda\le1$

But I don't know where is it come from ? Unfortunately , I can't understand it. I searched in the internet many times but it didn't help to me. If someone explain this expression is helpful. 

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    Perhaps the Answer posted by C. Falcon will help. Saying a function of one variable is *convex* essentially amounts to saying the region *above* the graph of the function is convex. Sometimes we say the graph is "concave up" rather than calling the function convex, but the notions are equivalent. – 2017-01-14
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    Okay , but I have still problem . What is the meaning of "set of barycentric combinations of $A$ and $B$ "? – 2017-01-14
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    A different definition is that a function is convex iff its epigraph is convex. This is a visually intuitive definition, and it explains why we use the same word "convex" for both sets and functions. – 2017-01-14
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    @hardmath A [convex curve](https://en.wikipedia.org/wiki/Convex_curve) is commonly defined as one that stays on the same side of any of its tangents. In this sense, both graphs $y=x^2$ and $y=-x^2$ are convex curves, even though the former is defined by a [convex function](https://en.wikipedia.org/wiki/Convex_function) while the latter by a concave function. The geometrical definition of a convex function is that the set of points *above* its graph $y \ge f(x)$ (a.k.a. its [epigraph](https://en.wikipedia.org/wiki/Epigraph_(mathematics))) is convex . – 2017-01-14
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    @S.H.W If you look at the very first picture on the wikipedia page for [convex functions](https://en.wikipedia.org/wiki/Convex_function) you'll see what the inequality means geometrically (except they use $t$ instead of $\lambda$). – 2017-01-14
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    @dxiv Yes , you're right. I read it but I'm really confuse about notions – 2017-01-14

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The given inequality means that whenever you take two points on the graph of $f$, then the segment joining them is above the graph itself.

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Edit. For the proof, it suffices to notice that for all $A,B\in\mathbb{R}^2$, one has: $$[AB]=\{(1-t)A+tB;t\in[0,1]\}.$$ This is essentially the definition of $[AB]$, which is the set of barycentric combinations of $A$ and $B$.

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    I know it , but can you show it mathematically ? – 2017-01-14
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    Are you familiar with parametrizing line segments starting from a point $a$ to a point $b$? – 2017-01-14
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    @Andrew Whelan Is your purpose slope of it ? – 2017-01-14
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    @C. Falcon Can you explain "set of barycentric combinations of $A$ and $B$ "? – 2017-01-14
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    @S.H.W A barycentric combination of points is a finite linear combination of points whose weights are positive and sum up to $1$. Intuitively, if $A$ has weight $\lambda\geqslant 0$ and $B$ weight $\mu\geqslant 0$, with $\lambda+\mu=1$, then the barycenter of the system $\{(A,\lambda),(B,\mu)\}$ is its center of mass. When I say that $[AB]$ is the set of barycentric combinations of $A$ and $B$, I mean that any point in $[AB]$ is given as the center of mass of such a system. For example, $A$ is the barycentric combination of $A$ with weight $1$ and $B$ with weight $0$. – 2017-01-14
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    @C. Falcon I know center of mass in physics but it seems that notion here is different from physics . Am I right ? – 2017-01-14
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    @S.H.W This is really the same thing. Perhaps in Physics you were taught that the center of mass of $(A,m_A)$ and $(B,m_B)$ is the unique point $G$ such that: $$m_A\overrightarrow{GA}+m_B\overrightarrow{GB}=\overrightarrow{0}.$$ Can you see the analogy with $(1-t)A+tB$? – 2017-01-14
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    Yes , but I'm not familiar with Barycentric Coordinates . Can you provide a resource for it ? – 2017-01-14
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    @S.H.W All the references I am familiar with are in French, however I believe the translation of *Géométrie* by M. Berger would be fine. See chapter $3$, part $4$ of the volume $1$. – 2017-01-14
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    Okay , Thanks a lot. – 2017-01-14
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Copied from my answer to the question What is the intuition behind the mathematical definition of convexity?; that is essentially a duplicate, so I have just flagged it as such.

The idea of convexity is is applicable in the first place to shapes or their surfaces and means bulging with no dents. This concept can be applied when the shape is a set of points in a space for which we can define a “dent”; Euclidean spaces will do. It can also apply to part of the surface with no dents.

We can think of a dent as a place where you can draw a straight line segment joining two points in the set but leaving the set somewhere along that segment. If the set is “well-behaved” and has a surface, such a segment leaves the set at some point and re-enters it another, there is a subsegment joining points on the surface. In this case, we may define convex by saying all points on such segments lie in the set.

Derived from that, a function is described as convex when the set of points above (or maybe below) of its graph is convex. Note that a function may be convex upwards or downwards, with the unqualified form meaning “convex downwards”. Further, as in your case, we call a function convex on an interval if the set of points above the graph with $x$ in that interval is convex.

The formulation with $λ$ and $1-λ$ formalises the above definition for the case of a function, that all points on a segment between points on the graph lie in the set: one side gives the value of the function $λ$ of the way along $[x_1,x_2]$, the other, the point that far along the segment joining two points on the line; the inequality says the point on the segment is above the graph, i.e. in the set.