Seems ugly, But cute

[Thamim Zahin]

Find all positive real solutions $(x_1, x_2, x_3, x_4, x_5, a)$ to the following set of equations:

$$ \displaystyle\sum_{k=1}^{5} kx_k=a $$

$$ \displaystyle\sum_{k=1}^{5} kx^3_k=a^2 $$

$$ \displaystyle\sum_{k=1}^{5} kx^5_k=a^3 $$

[Thanic Nur Samin]

Apply cauchy-schwarz inequality.

$$\displaystyle \left(\sum_{k=1}^5 kx_k\right)\left(\sum_{k=1}^5 kx_k^5\right)\ge \left(\sum_{k=1}^5 kx_k^3\right)^2$$

$$\Rightarrow a\times a^3\ge (a^2)^2$$

So this is the equality case of the inequality. So we get that all $x_k$'s are equal. Let them be $x$.

So, we get, $15x=a$, $15x^3=a^2$ and $15x^5=a^3$. Dividing the second by third, we get $a=x^2$, and substituting this yields $x=15$ and $a=225$ which works for all equations.

[Thamim Zahin]

It took me more than 9 hours to do. Now the solution:

If we multiple the first equation and the the third equation, we get:

$(x_1+2x_2+3x_3+4x_4+5x_5)(x_1+2x_2+3x_3+4x_4+5x_5)$ $=$


$x^6_1+2x_2x^5_1+3x_3 x^5_1+4x_4 x^5_1+5x_5 x^5_1+$

$2x_1 x^5_2+4x^6_2+6x_3 x^5_2+8x_4 x^5_2+10x_5 x^5_2$ $+$

$3x_1 x^5_3+6x_2 x^5_3+9x^6_3+12x_4 x^5_3+15x_5 x^5_3+$

$4x_1 x^5_4+8x_2 x^5_4+12x_3 x^5_4+16x^6_4+20x_5 x^5_4+$

$5x_1 x^5_5+10x_2 x^5_5$ $+$ $15x^5_3 x_5+20x^5_4 x_5+25x^6_5 =a^4$

And by squaring the 2nd equation and simplify, we get,

$(x^3_1+x^3_2+x^3_3+x^3_4+x^3_5)^2=$

$x^6_1+4x^6_2+9x^6_3+16x^6_4+25x^6_5+$

$4x^3_1 x^3_2+6x^3_1 x^3_3+$ $8x^3_1 x^3_4+10x^3_1 x^3_5+$

$12x^3_2 x^3_3+16x^3_2 x^3_4+$ $20x^3_2 x^3_5+$

$24x^3_3 x^3_4+30x^3_3 x^3_5+$ $ 40x^3_4 x^3_5=a^4$


By combining the both equation and simplifying, we get,

$2x_2x^5_1+3x_3 x^5_1+4x_4 x^5_1+5x_5 x^5_1+$

$2x_1 x^5_2+6x_3 x^5_2+8x_4 x^5_2+10x_5 x^5_2$ $+$

$3x_1 x^5_3+6x_2 x^5_3+12x_4 x^5_3+15x_5 x^5_3+$

$4x_1 x^5_4+8x_2 x^5_4+12x_3 x^5_4+20x_5 x^5_4+$

$5x_1 x^5_5+10x_2 x^5_5$ $+$ $15x^5_3 x_5+20x^5_4 x_5$ $ =$

$4x^3_1 x^3_2+6x^3_1 x^3_3+$ $8x^3_1 x^3_4+10x^3_1 x^3_5+$

$12x^3_2 x^3_3+16x^3_2 x^3_4+$ $20x^3_2 x^3_5+$

$24x^3_3 x^3_4+30x^3_3 x^3_5+$ $ 40x^3_4 x^3_5$

Now by applying AM-GM, we can show that every 2 variable of LHS with same co-efficient is greater than or equal to one of the variable of RHS (That one where the sum of that two co-efficient is equal to the sum of the RHS variable's co-efficient). For example: $2x_1 x^5_2+2x_2x^5_1 \ge 4x^3_1 x^3_2$

It same of for the other 9(We have total 10 pair including this)

But, all of the inequalities sum is equal. That mans all the inequalities are equal. For rxample $2x_1 x^5_2+2x_2x^5_1 = 4x^3_1 x^3_2$.
It is also true for the other 9.

From this we can easily show that all $x_k$'s are equal.

By this we get, $15x=a, 15x^3=a^2, 15x^5=a^3$

Dividing the 2nd by 3rd, we get $a=x^2$

Putting them in the first equation yield us that $(x,a)=(15,225)$

Which works for all the equation.

$[Done]$

[ahmedittihad]

Thamim really outdone himself this time.

[Thanic Nur Samin]

Thamim really outdone himself this time.

In fact, a proof of cauchy-schwarz inequality relies on the expansion of the product, and in that sense while Thamim's solution might look lengthy, they are actually same in the core.

Of course, I like the vector product proof of cauchy-schwarz inequality better

[ahmedittihad]

His proof isn't much ugly. I'm more amazed at his patience of latexing.