Let f be a twice differentiable function such that $f''>0$ for every x prove that $f(x+1)+f(x-1) > 2 f(x)$ Prove that :
$$f(x+1)+f(x-1) > 2 f(x)$$
I know $f''(x)>0$ for every x therefore $f'(x)$ is stricly increasing then I know I gotta use the mean value theorem in some way but I cant reach this relation best I got is
I used mean value formula for $f'(x)$ and $f'(x+2)$ and used the fact the derivative is increasing to get $$f(x+1) - f(x-1) < f(x+3) -f(x+1)$$ Then I used that to get the following after switching sides
$$2f(x)< f(x+2) + f(x-2)$$ which is not what I was asked to prove and I dont know how to continue from here
 A: Method 1.  By the MVT there exist $y\in (x-1,x)$ and $z
\in (x,x+1)$ with $f'(y)=\dfrac {f(x)-f(x-1)}{x-(x-1)}=f(x)-f(x-1)$ and $f'(z)=\dfrac {f(x+1)-f(x)}{(x+1)-x}=f(x+1)-f(x).$
Now $y<z$ because $y<x<z.$ So by the MVT there exists $w\in (y,z)$ with $$0<f''(w)=\frac {f'(z)-f'(y)}{z-y}=$$ $$=\frac {(f(x+1)-f(x))-(f(x)-f(x-1))}{z-y}=$$ $$=\frac {f(x+1)+f(x-1)-2f(x)}{z-y}$$ which implies $0<f(x+1)+f(x-1)-2f(x)$ because $0<z-y.$
Method 2. By a deeper theorem on remainders of power series, if $r\ne 0$ then $f(x+r)=f(x)+rf'(x)+r^2f''(s_{x,r})/2$ for some $s_{x,r}$ that lies strictly between $x$ and $x+r$. Applying this with $r=1$ and again with $r=-1,$ we have $$f(x+1)=f(x)+f'(x)+f''(s_{x,1})/2$$ $$ f(x-1)=f(x)-f'(x)+f''(s_{x,-1})/2$$ Adding these gives $$f(x+1)+f(x-1)=2f(x)+f''(s_{x,1})/2+f''(s_{x,-1})/2>2f(x).$$
A: As suggested in the comments, since $f'' >0$, the function will be strictly convex on its domain and, in particular, in the interval $[x-1, x+1]$. So, for every $\lambda \in (0,1)$,
$$
f((1-\lambda)(x-1) + \lambda(x+1)) < (1-\lambda) f(x-1) + \lambda f(x+1)
$$
If you take, in particular, $\lambda = \frac 12$, you get
$$
f(x) < \frac 12 f(x-1) + \frac 12 f(x+1),
$$
which is equivalent to the result you mention.
