Prove that $\sqrt[n]{a_1}-\sqrt[n]{a_2}+\sqrt[n]{a_3}-\dots-\sqrt[n]{a_{2n}}+\sqrt[n]{a_{2n+1}}<\sqrt[n]{a_1-a_2+a_3-\dots-a_{2n}+a_{2n+1}}$

contest-mathinequality

Question:

Prove that $\sqrt[n]{a_1}-\sqrt[n]{a_2}+\sqrt[n]{a_3}-\dots-\sqrt[n]{a_{2n}}+\sqrt[n]{a_{2n+1}}<\sqrt[n]{a_1-a_2+a_3-\dots-a_{2n}+a_{2n+1}}$ where $n$ and $\{a_i\}$ are reals and $n\geq2$ and $0<a_1<a_2<a_3<\dots<a_{2n+1}$.

This is a Math Olympiad question for $7$~$8$th-graders.

My attempt:

The restriction $0<a_1<a_2<a_3<\dots<a_{2n+1}$ makes the question really hard. The only inequalities that work are Jenson's and Karamata's. However, Jenson's does not work here as the weights ($1, -1, 1, -1, \dots$) are not all positive. Karamata's does not work here either as there are only $1$ sequence $\{a_i\}$.

I tried raising both sides to a power of $n$ to try and simplify the expression but it did not help.

$$
\sqrt[n]{a_1}-\sqrt[n]{a_2}+\sqrt[n]{a_3}-\dots-\sqrt[n]{a_{2n}}+\sqrt[n]{a_{2n+1}}<\sqrt[n]{a_1-a_2+a_3-\dots-a_{2n}+a_{2n+1}} \\
\iff (\sqrt[n]{a_1}-\sqrt[n]{a_2}+\sqrt[n]{a_3}-\dots-\sqrt[n]{a_{2n}}+\sqrt[n]{a_{2n+1}})^n<(\sqrt[n]{a_1-a_2+a_3-\dots-a_{2n}+a_{2n+1}})^n \\
\iff a_1-a_2+a_3-\dots-a_{2n}+a_{2n+1}+\text{*ugly expression*}<a_1-a_2+a_3-\dots-a_{2n}+a_{2n+1}
$$

How can I do this question? Any help will be appreciated.

Best Answer

To prove this using OP's ideas:

  1. Show that $f(x) = \sqrt[n](x)$ is a concave function.
  2. Show that $f(a_1) + f(a_3) < f(a_2) + f(a_1-a_2+a_3)$ for $a_1 < a_2 < a_3$.
    • This should be "obvious" from a visual graph. To formally prove it, use Jensen's, applied to a weighted average
    • Show that $f(a_2) = f(\frac{(a_2 - a_1)a_3 + (a_3 - a_2)a_1 } { a_3 - a_1}) > \frac{a_2 - a_1}{a_3 - a_1} f(a_3) + \frac{a_3 - a_2}{a_3-a_1} f(a_1). $.
    • Show that $a_1 < a_1 - a_2 + a_3 < a_3$.
    • Find a similar inequality for $f(a_1 - a_2 + a_3)$.
    • Add up these 2 inequalities.
  3. Recognize that we can induct on the number of terms $a_1, a_2, \ldots a_{2m+1}$.
    • Assume it is true for some $m$. For some increasing sequence $\{b_i\}_{i=1}^{2m+3}$, consider $a_1 = b_1 - b_2 + b_3$ and $a_i = b_{i+2}$ for $i = 2$ to $2m+1$.
    • Show that $\{a_i\}_{i=1}^{2m+1}$ is an increasing sequence so we can apply the induction hypothesis
    • Show that $\sum (-1)^{i+1} f(b_i) < f(b_1 - b_2 + b_3) - f(b_4) + \ldots f(b_{2m+3} ) < f( \sum(-1)^{i+1} b_i)$.
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