Question: A function f is defined to be convex on the closed interval $[1,5]$ if and only if $f(t+(1-t)5) \le tf(1) + (1-t)f(5)$ for any $t$ between $0$ and $1$ inclusive $(0\le t\le 1).$

Please help me prove this or at least where to start and what I am looking for at the end. I am lost.

I would really appreciate it!

Thanks!

Answer: 
$f(1)=(1-3)^2=4$,
$f(5)=(5-3)^2=4$, so the right hand side is $tf(1)+(1-t)f(5)=4t+(1-t)4=4$. The left hand side is $f(t+(1-t)5)=f(5-4t)=(5-4t-3)^2=(2-4t)^2=4(1-t)^2$.

Note that $0\leq t\leq 1$, so $0\leq 1-t\leq 1$, $0\leq(1-t)^2\leq 1$. 
Multiply by 4 to obtain

$4(1-t)^2\leq 4$. Q.E.D.
In this case,
the function has a second derivative
and $f''(x) = 2 > 0$
so it is convex.