I need to prove that

$$\frac{k(k+1)}{2}\left(\frac{a_1^2}{k} + \frac{a_2^2}{k-1} + \ldots + \frac{a_k^2}{1}\right) \geq (a_1 + a_2 + \ldots + a_k)^2\;,$$ where $a_1, a_2, \dots, a_k$ is some set of reals.


Can I presume without the loss of generality that $a_1 \leq a_2 \leq \ldots \leq a_n$ ?

This is how far I got:

I used the formula $\left \langle a,b \right \rangle \leq |a||b|$:

$$\begin{align*}\left \langle a,1 \right \rangle &\leq |a||1|\\ (a_1 + a_2 + \ldots + a_k) &\leq \sqrt{(a_1^2 + a_2^2 + \ldots + a_k^2)}\sqrt{k} \end{align*}$$
Square it:
$$(a_1 + a_2 + \ldots + a_k)^2 \leq k(a_1^2 + a_2^2 + \ldots + a_k^2)$$ Now I have to prove that:
$$\frac{k(k+1)}{2}\left(\frac{a_1^2}{k} + \frac{a_2^2}{k-1} + \ldots + \frac{a_k^2}{1}\right) \geq k(a_1^2 + a_2^2 + ... + a_k^2)$$ But I'm not sure how. Any pointers?

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    $\begingroup$ You don't want to use an arbitrary intermediary step like that because the result is not true anymore. What you want is to interpret the very first inequality as one instance of Cauchy-Schwarz. Write down Cauchy-Schwarz not using the letter $a$ and match up terms to determine what you need to plug in. $\endgroup$ – Phira May 17 '12 at 10:27
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    $\begingroup$ Maybe the identity $\frac{k(k+1)}2 = \sum_{j=1}^k j$ will help. You have the numbers 1, ..., k already, and the factor $\frac{k(k+1)}2$. $\endgroup$ – Daan Michiels May 17 '12 at 10:28
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    $\begingroup$ You may not assume that the $a_i$ are non-decreasing, because they’re not treated identically: interchanging $a_1$ and $a_k$, for instance, changes $$\frac{a_1^2}k+\frac{a_k^2}1\;.$$ $\endgroup$ – Brian M. Scott May 17 '12 at 10:30
  • $\begingroup$ I wonder why (linear-algebra) tag. $\endgroup$ – Martin Sleziak Apr 29 '17 at 6:10

Try considering instead the vectors $(\frac{a_1}{1},...,\frac{a_k}{\sqrt{k}})$ and $(1,...,\sqrt{k})$.

  • $\begingroup$ I'm sorry, I don't quite follow $\endgroup$ – neoric May 17 '12 at 10:39
  • $\begingroup$ Try using these as the vectors a and b in ⟨a,b⟩≤|a||b| that you used. $\endgroup$ – Wonder May 17 '12 at 10:42
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    $\begingroup$ Well strictly speaking you have to use $(\frac{a_1}{\sqrt{k}},...,\frac{a_k}{1})$ and $(\sqrt{k}, ..., 1)$. $\endgroup$ – Wonder May 17 '12 at 10:45

You can simply use the inequality of quadratic and arithmetic mean for $k$ elements $\frac{a_1}k$, $k-1$ elements $\frac{a_2}{k-1}$ etc. For the inequality between quadratic and arithmetic mean see e.g. Jensen inequality and Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality at AoPS.

Arithmetic mean is $$a=\frac{a_1+\dots+a_k}{\frac{k(k+1)}2}.$$

Quadratic mean is $$q=\sqrt{\frac{\frac{a_1^2}k+\frac{a_2^2}{k-1}+\dots+a_k^2}{\frac{k(k+1)}2}}.$$

So from $q^2\ge a^2$ you get $$\frac{\frac{a_1^2}k+\frac{a_2^2}{k-1}+\dots+a_k^2}{\frac{k(k+1)}2} \ge \left(\frac{a_1+\dots+a_k}{\frac{k(k+1)}2}\right)^2$$ and $$\frac{k(k+1)}2 \left(\frac{a_1^2}k+\frac{a_2^2}{k+1}+\dots+a_k^2\right) \ge (a_1+\dots+a_k)^2.$$


I am giving a solution just by using Cauchy-Schwarz inequality (as proceed by the poster of this problem), namely $(a,b)\leq ||a||||b||$ where $a,b\in \mathbb R^{n}$

Take $n=\frac{k(k+1)}{2}$ and $a=b$ where $a$ is a $\frac{k(k+1)}{2}$ tuple vector with first $k$ entries are $\frac{a_{1}}{k}$, second $(k-1)$ entries are $\frac{a_{2}}{k-1}$, next $(k-2)$ entries are $\frac{a_{3}}{k-2}$....likewise... last entry is $\frac{a_{k}}{1}$.

  • $\begingroup$ There is a mistake take $b=$ all 1's $\endgroup$ – users31526 May 17 '12 at 16:48
  • $\begingroup$ With this mistake one can prove another inequality.... $\endgroup$ – users31526 May 17 '12 at 16:54

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