Crazy Inequalities 2 + 400 followers problem

The Cauchy-Schwarz inequality for $(a_{1}, a_{2}, \ldots, a_{n}), (b_{1}, b_{2}, \ldots, b_{n}) \in \mathbb{R}^{n}$ only achieves equality for when both ordered tuples (vectors) are linearly dependant.

Taking $a_{k} = \sqrt{x_{k}}$ and $b_{k} = \frac{k^{3}}{x_{k}^{2}\sqrt{x_{k}}}$ , in order for the minimum to be achieved, we need

$\frac{A^{3}k^{3}}{x_{k}^{2}\sqrt{x_{k}}} = \sqrt{x_{k}} \longrightarrow x_{k} = Ak$

for some nonzero scalar $A$ .

Substituting $x_{k}$ into the given sum, we see

$\sum_{k=1}^{400} \frac{k^{3}}{x_{k}^{2}} = \frac{1}{A^{2}} \sum_{k=1}^{400} k = \frac{400 \times 401}{2A^{2}} = 1$

Hence, $A = \sqrt{\frac{400 \times 401}{2}}$ . Finally, we subsitute in $x_{k}$ again:

$\sum_{k=1}^{400} x_{k} = A\frac{400 \times 401}{2} = A^{3}$