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.
As per my comment, we must show
$$f(tx + (1 - t)y) \le tf(x) + (1 - t)f(y)$$
for $t \in [0, 1]$ and $x, y \in [1, 5]$. Actually, $f(x) = (x - 3)^2$ is convex everywhere, so you'll find that we don't really need the assumption that $x, y \in [1, 5]$.

We have,

\begin{align\*}
&tf(x) + (1 - t)f(y) - f(tx + (1- t)y) \\
= ~ &t(x - 3)^2 + (1- t)(y - 3)^2 - (tx + (1 - t)y - 3)^2 \\
= ~ &t(x - 3)^2 + (1- t)(y - 3)^2 - (t(x - 3) + (1 - t)(y - 3))^2 \\
= ~ &t(x - 3)^2 + (1- t)(y - 3)^2 - t^2(x - 3)^2 - (1 - t)^2(y - 3)^2 - 2t(1 - t)(x - 3)(y - 3) \\
= ~ &t(1 - t)(x - 3)^2 + t(1- t)(y - 3)^2 - 2t(1 - t)(x - 3)(y - 3) \\
= ~ &t(1 - t)[(x - 3)^2 + (y - 3)^2 - 2(x - 3)(y - 3)] \\
= ~ &t(1 - t)[(x - 3) - (y - 3)]^2 \\
= ~ &t(1 - t)(x - y)^2 \ge 0,
\end{align\*}
which is what we wanted to prove.