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I was trying to prove for two positive integers $i$ and $j$ and a natural number $n > 1$ whether the following statement holds:

If $a_{0}^{n} + a_{1}^{n} + \ldots + a_{i}^{n} > b_{0}^{n} + b_{1}^{n} + \ldots + b_{j}^{n}$ is true, then $a_{0}^{n + 1} + a_{1}^{n + 1} + \ldots + a_{i}^{n + 1} > b_{0}^{n + 1} + b_{1}^{n + 1} + \ldots + b_{j}^{n + 1}$ is true as well.

Where every $a_{k} \geq 0$ for $0 \leq k \leq i$ and every $b_{k} \geq 0$ for $0 \leq k \leq j$. I was fighting against this problem even when $i = j$ but I have no receive any further step forward.

Any help will be highly appreciated. Thanks in Advance.

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  • $\begingroup$ If we could do this by induction on $i$ and $j$, then we can do it by induction on $i$ assuming that $i = j$ since we can make arbitrarily $a_{k} = 0$ or $b_{k} = 0$ to correspond the equality $i = j$ $\endgroup$ – Frank Vega Mar 24 at 13:33
  • $\begingroup$ For $i = 1$ is simple to prove that $a_{0}^{n} > b_{0}^{n}$ implies $a_{0}^{n + 1} > b_{0}^{n + 1}$, but when we assume for some $i = k$ and try to prove $i = k + 1$ things get be harder... $\endgroup$ – Frank Vega Mar 24 at 13:36
  • $\begingroup$ Suppose we assume for some $k$ and try to prove for $k + 1$ that $a_{0}^{n} + \ldots a_{k + 1}^{n} > b_{0}^{n} + \ldots + b_{k + 1}^{n}$ implies $a_{0}^{n + 1} + \ldots a_{k + 1}^{n + 1} > b_{0}^{n + 1} + \ldots + b_{k + 1}^{n + 1}$, then if we remove $a_{k + 1}$ and $b_{k + 1}$ the statement is holds for $k$, but what happens when we add again $a_{k + 1}$ and $b_{k + 1}$? $\endgroup$ – Frank Vega Mar 24 at 13:42
  • $\begingroup$ if $a_{k + 1} \geq b_{k + 1}$, then the result is trivial, but @PeldePinda what happens when $a_{k + 1} < b_{k + 1}$... $\endgroup$ – Frank Vega Mar 24 at 13:53
  • $\begingroup$ Ahhhh, I already know how to solve it, when we pick $a_{k'}$ and $b_{k'}$ into $\{a_{0}, \ldots, a_{k + 1}\}$ and $\{b_{0}, \ldots, b_{k + 1}\}$, then I choose them such that $a_{k'} \geq b_{k'}$. In case there is no $k'$, then $a_{0}^{n} + \ldots a_{k}^{n} > b_{0}^{n} + \ldots b_{k}^{n}$ does not comply... $\endgroup$ – Frank Vega Mar 24 at 14:00
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The statement is already wrong for sums of two powers: $$ 10^2 + 10^2 = 200 > 170 = 13^2 + 1^2 \, , $$ but $$ 10^3 + 10^3 = 2000 < 2198 = 13^3 + 1^3 \, . $$

Graphically: The green curves are the circles $$ x^2 + y^2 = 170 \, , \quad x^2 + y^2 = 200 \, , $$ and the red curves are the superellipses $$ x^3 + y^3 = 2000 \, , \quad x^3 + y^3 = 2198 \, . $$ The point $A=(10, 10)$ lies on the “larger” circle and the “smaller” superellipse, whereas $B=(13, 1)$ lies on the “smaller” circle and the “larger” superellipse:

enter image description here

Remark: Your induction proof does not work because after removing $a_{k'}^n$ from the left sum and $b_{k''}^n$ from the right sum (with $a_{k'} > b_{k''}$) the remaining sums do not necessarily satisfy the induction hypothesis. In the above example you would remove $10^2$ from the left sum and $1^2$ from the right sum, and the remaining sums are $$ 10^2 \not\gt 13^2 \, . $$

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  • $\begingroup$ Thank very much!!! $\endgroup$ – Frank Vega Mar 25 at 13:55
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Within the parameters of the question as posed, it is possible to construct a trivial counterexample. You mention the possibility that $i=j$ but do not make that a requirement of the question. You require only that $a_k,b_k>0$. Your assumption $n\in \mathbb N$ goes to a specific number, as assuming it true for all $n$ generally assumes the conclusion.

Let $i=10,\ j=5,\ n=2$. Let $a_k=1,\ b_{1,2,3,4}=1,\ b_5=2$.

$\sum a_k^n=\sum a_k^{n+1}=10$

$\sum b_k^n=8;\ \sum b_k^{n+1}=12$

You may be able to avoid such trivial counterexamples by putting more conditions on your variables.

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  • $\begingroup$ Your answer is correct!!! Math stackexchange does not let me to put two answers as correct!!! Thank you very much!!! $\endgroup$ – Frank Vega Mar 25 at 13:55

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