Prove the inequality $\frac{k(k+1)}{2}\left(\frac{a_1^2}{k} + \frac{a_2^2}{k-1} + \ldots + \frac{a_k^2}{1}\right) \geq (a_1 + a_2 + \ldots + a_k)^2$ I need to prove that
$$\frac{k(k+1)}{2}\left(\frac{a_1^2}{k} + \frac{a_2^2}{k-1} + \ldots + \frac{a_k^2}{1}\right) \geq (a_1 + a_2 + \ldots + a_k)^2\;,$$ where $a_1, a_2, \dots, a_k$ is some set of reals.  
Firstly: 
Can I presume without the loss of generality that $a_1 \leq a_2 \leq \ldots \leq a_n$ ?
This is how far I got: 
I used the formula $\left \langle a,b \right \rangle \leq |a||b|$:  
$$\begin{align*}\left \langle a,1 \right \rangle &\leq |a||1|\\
(a_1 + a_2 + \ldots + a_k) &\leq \sqrt{(a_1^2 + a_2^2 + \ldots + a_k^2)}\sqrt{k}
\end{align*}$$
Square it:
$$(a_1 + a_2 + \ldots + a_k)^2 \leq k(a_1^2 + a_2^2 + \ldots + a_k^2)$$
Now I have to prove that:
$$\frac{k(k+1)}{2}\left(\frac{a_1^2}{k} + \frac{a_2^2}{k-1} + \ldots + \frac{a_k^2}{1}\right) \geq k(a_1^2 + a_2^2 + ... + a_k^2)$$
But I'm not sure how. Any pointers?

You can simply use the inequality of quadratic and arithmetic mean for $k$ elements $\frac{a_1}k$, $k-1$ elements $\frac{a_2}{k-1}$ etc. For the inequality between quadratic and arithmetic mean 
see e.g. Jensen inequality
and Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality at AoPS.
Arithmetic mean is $$a=\frac{a_1+\dots+a_k}{\frac{k(k+1)}2}.$$
Quadratic mean is $$q=\sqrt{\frac{\frac{a_1^2}k+\frac{a_2^2}{k-1}+\dots+a_k^2}{\frac{k(k+1)}2}}.$$
So from $q^2\ge a^2$ you get
$$\frac{\frac{a_1^2}k+\frac{a_2^2}{k-1}+\dots+a_k^2}{\frac{k(k+1)}2} \ge \left(\frac{a_1+\dots+a_k}{\frac{k(k+1)}2}\right)^2$$
and
$$\frac{k(k+1)}2 \left(\frac{a_1^2}k+\frac{a_2^2}{k+1}+\dots+a_k^2\right) \ge (a_1+\dots+a_k)^2.$$ 
