<s>[INST] 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?
 [/INST] I am giving a solution just by using Cauchy-Schwarz inequality (as proceed by the poster of this problem), namely $(a,b)\leq ||a||||b||$ where $a,b\in \mathbb R^{n}$
Take $n=\frac{k(k+1)}{2}$ and $a=b$ where $a$ is a $\frac{k(k+1)}{2}$ tuple vector with first $k$ entries are $\frac{a_{1}}{k}$, second $(k-1)$ entries are $\frac{a_{2}}{k-1}$, next $(k-2)$ entries are $\frac{a_{3}}{k-2}$....likewise... last entry is $\frac{a_{k}}{1}$.
</s>