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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.

Firstly:

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?

  • 0
    I wonder why ([tag:linear-algebra]) tag.2017-04-29

3 Answers 3

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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 $\frac{a_{k}}{1}$.

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    Wit$h$ this mistake one can prove another inequality....2012-05-17
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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 $\frac{k(k+1)}2 \left(\frac{a_1^2}k+\frac{a_2^2}{k+1}+\dots+a_k^2\right) \ge (a_1+\dots+a_k)^2.$

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

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    Well strictly speaking you have to use $(\frac{a_1}{\sqrt{k}},...,\frac{a_k}{1})$ and $(\sqrt{k}, ..., 1)$.2012-05-17