For positive real numbers solve the system of equations $$ \begin{cases} \begin{align} x_1 + x_2 + · · · + x_n = \frac{1}{4} \\ \frac{1}{x_1} + \frac{4}{x_2} +· · ·+ \frac{n^2}{x_n} = n^2(n+1)^2\\ \end{align} \end{cases} $$
I used Cauchy Schwarz for roots ($\sqrt{(a_1 + a_2+· · ·+a_n)(b_1 + b_2+· · ·+b_n)} \geq \sqrt{a_1b_1}+\sqrt{a_2b_2} + · · ·+ \sqrt{a_nb_n}$)
This yields that equality must occur. I don't know how to finish this. I can't find the exact name of this inequality and hence can't say when equality occurs in the inequality.
Obviously $a_1=a_2=· · ·=a_n$ and $b_1=b_2=· · ·=b_n$ would satisfy.
This yields that there are no solutions.
I am unsure hence I mark the question as proof verification as well.
Also if someone knows the name of the inequality it would be appreciated, in my language we call it just cauchy schwarz for roots, CS for fractions, classic CS etc.
Thanks in advance!
Titu's Lemma might be what you're looking for. Applying it yields: \begin{align*} \frac{1}{x_1} + \frac{4}{x_2} + \cdots + \frac{n^2}{x_n} &\geq \frac{(1 + 2 + \cdots + n)^2}{x_1 + x_2 + \cdots + x_n}\\ &= \frac{\frac{n^2(n+1)^2}{4}}{\frac{1}{4}} \\ &= n^2(n+1)^2 \end{align*} This lemma is simply an application of CS Inequality, with $a_i = \frac{k}{\sqrt{x_k}}$ and $b_i = \sqrt{x_k}$. Now CS Inequality holds iff: \begin{align*} \frac{a_1}{b_1} = \frac{a_2}{b_2} = \cdots = \frac{a_n}{b_n} \end{align*} In other words: $$ \frac{1}{x_1} = \frac{2}{x_2} = \cdots = \frac{n}{x_n} $$ Can you continue?