Find all real $a$ that satisfy the following equations.

Question

$x_1, x_2, x_3, x_4 ,x_5$ are all real and non-negative. The equations are,

$\sum _{k=1}^5\left(k\cdot x_k\right)=a$

$\sum _{k=1}^5\left(k^3\cdot x_k\right)=a^2$

$\sum _{k=1}^5\left(k^5\cdot x_k\right)=a^3$

I am totally stuck! I have no idea how to do it. I can only go this far:

Answer 1

By C-S $$a^4=\sum_{k=1}^nkx_k\sum_{k=1}^5k^5x_k\geq\left(\sum_{k=1}^5k^3x_k\right)^2=a^4,$$ The equality occurs for $$(\sqrt{x_1},\sqrt{2x_2},...\sqrt{5x_5})||(\sqrt{x_1},\sqrt{32x_2},...,\sqrt{5^5x_5}),$$ which says $$x_2=x_3=x_4=x_5=0.$$ Can you end it now?

Answer 2

As @Michael Rozenberg already said: $$ a^4=a\cdot a^3=\sum_{i=1}^5\sum_{j=1}^5ij^5\cdot x_ix_j =\left(a^2\right)^2=\sum_{i=1}^5\sum_{j=1}^5i^3j^3\cdot x_ix_j $$ Comparing the coefficients at $x_ix_j=x_jx_i$, we see that they are $ij^5+i^5j$ on the LHS and $2i^3j^3$ on the RHS, and we indeed have $ij^5+i^5j \ge 2i^3j^3 \Leftrightarrow j^4+i^4 \ge 2i^2j^2 \Leftrightarrow \left(i^2-j^2\right)^2\ge 0$. So, we have equality if and only if $i=j$. Therefore, to match the above equation, it must be $$ x_ix_j=0 \text{ for all } i \ne j $$ In other words: At most one of the variables can be non-zero (clearly, $x_1=x_2=x_3=x_4=x_5=0$ is a solution if and only if $a=0$). Now, if $k\in \{1,2,3,4,5\}$ with $x_k \ne 0$ and $kx_k=a$, $k^3x_k=a^2$ and $k^5x_k=a^3$, we see that $a=k^2$ and so $x_k=k$. Therefore, we have $6$ solutions in total: $$\begin{align} x &= (0,0,0,0,0) \text{ for } a=0\text{,}\\ x &= (1,0,0,0,0) \text{ for } a=1\text{,}\\ x &= (0,2,0,0,0) \text{ for } a=4\text{,}\\ x &= (0,0,3,0,0) \text{ for } a=9\text{,}\\ x &= (0,0,0,4,0) \text{ for } a=16\text{,}\\ \text{and }x &= (0,0,0,0,5) \text{ for } a=25\text{.}\\ \end{align}$$

Answer 3

The solutions are: x1 = 1/2 (a^3 - 5 a^2 + 6 a - 8 x4 - 30 x5), x2 = 1/2 (-a^3 + 4 a^2 - 3 a + 12 x4 + 40 x5), x3 = 1/6 (a^3 - 3 a^2 + 2 a - 24 x4 - 60 x5) x4 independent x5 independent

and make sure that x1>=0, x2>=0, x3>=0, x4>=0 and x5>=0