# Inequality sum of exponentiation

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.

• 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 $i = j$ – Frank Vega Mar 24 at 13:33
• 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... – Frank Vega Mar 24 at 13:36
• 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 $b_{k + 1}$? – Frank Vega Mar 24 at 13:42
• if $a_{k + 1} \geq b_{k + 1}$, then the result is trivial, but @PeldePinda what happens when $a_{k + 1} < b_{k + 1}$... – Frank Vega Mar 24 at 13:53
• 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... – Frank Vega Mar 24 at 14:00

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:

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 $$10^2 \not\gt 13^2 \, .$$

• Thank very much!!! – Frank Vega Mar 25 at 13:55

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.

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