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Let $x_1, \cdots, x_n \geq 0$ be real nonnegative numbers satisfying $x_1 \leq x_2 \leq \cdots \leq x_n$. For all integer $1 \leq m \leq n$, let

$$ x_1^2 + \cdots + x_m^2 \geq m^2.$$

Show that $x_1 + \cdots + x_n \geq \sqrt{1} + \sqrt{3} + \cdots \sqrt{2n-1}$.

My intuition says that the best way to solve this problem will use QM-AM-GM, but I haven't been able to find anything. I did notice that the equality case occurs when $x_1 = \sqrt{1}$, $x_2 = \sqrt{3}$, ..., $x_n = \sqrt{2n-1}$. This follows from the formula $1 + 3 + 5 + \cdots + (2n-1) = n^2$.

Also, I was thinking that it could be easier to prove the slightly stronger hypothesis $x_1 + \cdots + x_m \geq \sqrt{1} + \sqrt{3} + \cdots + \sqrt{2m-1}$ by induction, but I haven't been able to get any results with this idea either.

Can anybody give me hints to point me in the right direction for solving this problem?

Michael Rozenberg
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aras
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3 Answers3

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Hints:
1. Argue it is enough to consider the cases with $\displaystyle \sum_1^n x_i^2 = n^2$.
2. Use Karamata's inequality with the concave function $t \mapsto \sqrt{t}$, after observing that $$(x_1^2, x_2^2, x_3^2, \dots x_n^2) \prec (1, 3, 5, \dots 2n-1)$$

Macavity
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  • This is a nice and simple solution, but it feels like using Karmata is like nuking a mosquito. Do you think there is a more elementary proof? – aras Dec 01 '16 at 11:54
  • @aras another argument using the linear objective function and convex domains comes easily to mind, though not any argument with QM-AM-GM or something like that. Have to go now though, will get back later if something along more basic inequalities strikes. – Macavity Dec 01 '16 at 11:58
  • OK, thanks anyway. – aras Dec 01 '16 at 12:01
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Here is another way. The constraints (after some work), boil down to $x_1\in [1, x_2], x_2\in [\max(\sqrt3, x_1), x_3], x_3\in[\max(\sqrt5, x_2), x_4], \dots x_n \in [\max(\sqrt{2n-1},x_{n-1}), \infty)$

Clearly the linear function gets minimised when the variables take the lower boundaries of their respective intervals. So we have the minimum when $x_i=\sqrt{2i-1}$ and the inequality is done.

Macavity
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1

I suppose I'm a little late to the party but here is a neat proof using Abel's summation formula: say $s_i=a_1^2+a_2^2+...+a_i^2$ and $s_0=0.$

Rewrite $$\sum_{i=1}^n a_i=\sum_{i=0}^{n-1}(s_{i+1}-s_i)\frac{1}{a_{i+1}}=\sum_{i=1}^{n-1} s_{i}(\frac{1}{a_i}-\frac{1}{a_{i+1}})+\frac{s_n}{a_n}.$$

Using the hypothesis, this is at least $$\sum_{i=1}^{n-1}i^2(\frac{1}{a_{i+1}}-\frac{1}{a_i})+\frac{n^2}{a_n}$$

And by applying again Abel's summation formula (in the other direction if you wish) we see the latter is equal to $$\sum_{i=1}^{n}(2i-1)\frac{1}{a_i}\overset{C-S}\geq \frac{(\sum_{i=1}^n \sqrt{2i-1})^2}{\sum_{i=1}^na_i}.$$

Thus, we've shown $$(a_1+a_2+...+a_n)\geq \frac{(\sqrt{1}+\sqrt{3}+...+\sqrt{2n-1})^2}{a_1+a_2+...+a_n}$$ and the conclusion follows.

ham_ham01
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