how to build a function that holds some condition I'm not a mathematician but I have a question. I have some conditions that I want to create a function which holds that condition, Is there any way to build a mathematical function to satisfy this condition.The condition I need the function to hold is 
$$
\frac{f(x + 2) + f(x)}2 \ge f(x + 1)
$$
I think I asked my question wrong,
I have a set of random and strickly increasing numbers, now I want to use some mathematical way to make them appear convex just like funxtion x^2, so I needed the function to satisfy the condition I noted above, but the real problem is that I want to make thease numbers to look convex, so is there anything I can use??
Thank you all
 A: $f(x) = x^2$ will do the job.
\begin{align}
\frac{f(x+2)+f(x)}{2} &= \frac{(x+2)^2 + x^2}{2} \\
 &= \frac{x^2 + 4x + 4 + x^2}{2} \\
 &= x^2 + 2x + 2  \\
 &> x^2 + 2x + 1  \\
 &= (x+1)^2  \\
 &= f(x+1).  \\
\end{align}
A: This condition is satisfied by any convex function, i.e., a function satisfying
$$
f\big(tx+(1-t)y\big)\le tf(x)+(1-t)f(y),
$$
for all $t\in (0,1)$.
If $f$ is twice differentiable, then it is convex if and only if 
$$
f''(x)\ge 0,
$$
for all $x$.
HENCE, try a function with non-negative 2nd derivative.
A: Your condition says that the function's value at $x+1$ is less than the average of its values at $x$ and $x+2$. In other words, at $x+1$, the function value must be below the "straight line"  values between $x$ and $x+2$. 
Functions with this sort of property are called "convex". Basically, this just means that the graph of the function looks "hollow" when viewed from above.
Many functions have this property. The simplest interesting one I can think of is $f(x) = x^2$.
A: There are many functions for which this condition holds, in general for function $f(x)$, and $0\leq\theta\leq 1$,  if:
$$\theta f(z)+(1-\theta) f(y)\geq  f(\theta z+(1-\theta)y)$$ then the function is called convex.  
In your case, $\theta=1/2$, $z=x+1$, $z=x$.
You can find a list of convex functions here.
