If $f''(x)\geq 0$ on $(a, b)$ then show that $f(\alpha x+(1-\alpha)y)\leq \alpha f(x)+(1-\alpha)f(y).$ If $f(x)$ be a function such that $f''(x)\geq 0$ on $(a, b)$ then applyint MVT show that for all $x,y\in (a, b)$ and for all $\alpha \in [0,1]$
$$f(\alpha x+(1-\alpha)y)\leq \alpha f(x)+(1-\alpha)f(y).$$
I know that I have to use Lagrange's MVT, but I don't understand how to use it here. Please help. 
 A: For simplicity, let $x<y$. Clearly $x<\alpha x+(1-\alpha)y<y$. Applying Lagrange' MVT in $[x,\alpha x+(1-\alpha)y], [\alpha x+(1-\alpha)y,y]$ respectively, one has
\begin{eqnarray}
f(\alpha x+(1-\alpha)y)-f(x)&=&f'(c_1)(1-\alpha)(y-x), c_1\in(x,\alpha x+(1-\alpha)y), \\
f(y)-f(\alpha x+(1-\alpha)y)&=&f'(c_2)\alpha(y-x), c_2\in(\alpha x+(1-\alpha)y,y).
\end{eqnarray}
Cleary $c_1<c_2$. Since $f''(x)\ge 0$, one has that $f'(x)$ is non-decreasing and hence $f'(c_1)\le f'(c_2)$ or
$$ \frac{1}{1-\alpha}[f(\alpha x+(1-\alpha)y)-f(x)]\le\frac{1}{\alpha}[f(y)-f(\alpha x+(1-\alpha)y)].$$
Thus one has
$$ f(\alpha x+(1-\alpha)y)\le \alpha f(x)+(1-\alpha)f(y). $$
A: Fix $x,y\in[a,b]$ (let us suppose that $x<y$) and consider, for all $\alpha\in[0,1]$ :
$$\phi(\alpha)=\alpha f(x)+(1-\alpha)f(y)-f(\alpha x+(1-\alpha)y)$$
Now compute :
$$\phi'(\alpha)=f(x)-f(y)-(x-y)f'(\alpha x+(1-\alpha)y)$$
By MVT, there exists some $c\in(x,y)$ such that $f(x)-f(y)=(x-y)f'(c)$;
Thus :
$$\phi'(\alpha)=(x-y)(f'(c)-f'(\alpha x+(1-\alpha)y))$$
Let us denote :$$\alpha_0=\frac{y-c}{y-x}$$
Since $f'$ is increasing, we can conclude that $\phi$ is increasing for $x\in[0,\alpha_0]$ and decreasing for $x\in[\alpha_0,1]$.
But $\phi(0)=\phi(1)=0$, so that $\forall\alpha\in[0,1],\,\phi(\alpha)\ge0$, which completes the proof.
