Prove $\sum_{i=1}^m a_i < \sum_{j=1}^n b_j $ given $\sum_{i=1}^m a_i^2 = \sum_{j=1}^n b_j^2 $ Let $n>m>0$, $a_i>0$, $b_j>0$, and $\sum_{i=1}^m a_i^2 = \sum_{j=1}^n b_j^2.$
Can we claim that
$\sum_{i=1}^m a_i < \sum_{j=1}^n b_j, $
and how to prove this?
 A: Mr. Winther has already written a nice comment showing that the conjecture is indeed not true. I just want to elaborate more on that and explain why his choices of $a$ and $b$ are indeed appropriate for this problem.
Observe that the statement $\sum_{i=1}^m a_i^2 = \sum_{j=1}^n b_j^2$ is equivalent to $||a||_2=||b||_2$, where $$
||x||_2=\sqrt{\sum_{i=1}^N |x_i|^2}
$$
for $x=(x_1,x_2,\dots,x_N)\in \Bbb R^N$. The restriction $a_i,b_j>0$ ensure that 
$$
||a||_1:=\sum_{i=1}^m |a_i|=\sum_{i=1}^m a_i\ ,\quad ||b||_1:=\sum_{i=1}^n|b_i|=\sum_{i=1}^n b_i\ .
$$

There are some abuses of notation here since $a$ and $b$ are from different $\Bbb R^N$.

By Jensen's Inequality, the relation 
$$
\frac 1{\sqrt{N}}||x||_1\le ||x||_2 \le ||x||_1
$$ 
holds. Suppose that $||x||_2=1$, the the first inequality becomes an equality when 
$$x=\left(\frac 1{\sqrt{N}},\frac 1{\sqrt{N}},\dots,\frac 1{\sqrt{N}} \right)
$$ 
while the second one holds when $x_i=1$ for exactly one $i\in\{1,2,\dots,N\}$ and $x_i=0$ everywhere else.
Using
$$
\frac 1{\sqrt{m}}||a||_1\le ||a||_2=||b||_2\le ||b||_1 \ ,
$$
the strongest inequality we could come up with should be 

$\frac 1{\sqrt{m}}||a||_1\le||b||_1$, which is the same as $\frac 1{\sqrt{m}}\sum_{i=1}^m a_i \le \sum_{j=1}^nb_i $.

The restriction that $a_i,b_j\ne 0$ is not very important here, since the norm function $||\cdot||$ varies continuously. Knowing this, it is not hard to come up with a counter example to your conjecture: 
Fix $||a||_2=||b||_2=1$, we try to make $||a||_1$ as large as possible by letting $$a=(1/\sqrt{m},1/\sqrt{m},\ldots,1/\sqrt{m})$$
and 
$$
b=(1,0,0,\dots,0)\ .
$$
To meet the restriction that $b_j>0$ we vary $b$ a little so that the other entries are not $0$. The exact amount can be computed if necessary.
