2
$\begingroup$

It is known that the arithmetic mean of a list of non-negative real numbers is less than or equal to the quadratic mean (root mean square) of the same list: $$\frac{x_1+x_2+\cdots+x_n}{n} \le \sqrt{\frac{x_1^2+x_2^2+\cdots+x_n^2}{n}}$$ (More about mean inequalities)

My question is that given the number $n$ (number of elements of the list) and the quadratic mean ($QM$) of that list can we find lower bound for the arithmetic mean ($AM$) of that list?

More precisely, what is the greatest lower bound for the $AM$ that we can find? For example, it is obvious that $AM \ge 0$. Also, it is easy to show that $AM \ge \dfrac{QM}{\sqrt{n}}$: $$(x_1+x_2+\cdots+x_n)^2 \ge x_1^2+x_2^2+\cdots+x_n^2 \quad \Rightarrow\\ x_1+x_2+\cdots+x_n \ge \sqrt{x_1^2+x_2^2+\cdots+x_n^2} \quad \Rightarrow\\ \frac{x_1+x_2+\cdots+x_n}{n} \ge \frac{1}{\sqrt{n}} \sqrt{\frac{x_1^2+x_2^2+\cdots+x_n^2}{n}} \quad \Rightarrow\\ AM \ge \frac{QM}{\sqrt{n}}.$$ Is there some better (greater) lower bound for the $AM$ if we know $n$ and $QM$?

$\endgroup$
2
$\begingroup$

The bound $AM \ge \frac{QM}{\sqrt{n}}$ is tight. The equality is achieved, for instance, when one of $x_i$’s equals $1$ and the remaining equal $0$.

$\endgroup$
1
+50
$\begingroup$

$\def\vec{\boldsymbol}\def\R{\mathbb{R}}$Usually by saying an inequality is tight, it means that some particular constant in it cannot be improved. E.g. in @AlexRavsky's answer, “$\text{AM} \geqslant \dfrac{\text{QM}}{\sqrt{n}}$ is tight” means that the inequality is true and the constant $\dfrac{1}{\sqrt{n}}$ cannot be replaced by a larger one, so what they proved is the following proposition:

$$\min_{\substack{\vec{x} \in \R_{\geqslant 0}^n\\\vec{x} ≠ \vec{0}}} \frac{\|\vec{x}\|_1}{\|\vec{x}\|_2} = 1,$$

where $\|\vec{x}\|_a = \left(\sum\limits_{k = 1}^n |x_k|^a \right)^{\frac{1}{a}}$. This, however, does not exclude the possibility that there exists a non-linear function $f$ of QM such that $\text{AM} \geqslant f(\text{QM})$ and $f(t) \geqslant \dfrac{t}{\sqrt{n}}$ for $t \geqslant 0$.

The following reasoning deals with the general scenario, but the result coincides with the linear bound due to the fact that AM and QM are homogeneous polynomials of $x_1, \cdots, x_n$ of the same order.

Proposition: For any $a > 1$ and $t \geqslant 0$,$$ \min_{\substack{\vec{x} \in \R_{\geqslant 0}^n\\\|\vec{x}\|_a = t}} \|\vec{x}\|_1 = t, $$ so the best function $f_a: [0, +∞) → \R$ satisfying $\|\vec{x}\|_1 \geqslant f_a(\|\vec{x}\|_a)$ for all $x \in \R_{\geqslant 0}^n$ is $f_a(t) = t$.

Proof: A lemma is needed.

Lemma: $(x + y)^a \geqslant x^a + y^a$ for $x, y \geqslant 0$.

Proof: Define $g(t) = (t + 1)^a - t^a$ for $t \geqslant 0$. Since $g'(t) = a ((t + 1)^{a - 1} - t^{a - 1}) \geqslant 0$, then $g(t) \geqslant g(0) = 1$ for $t \geqslant 0$.

Now for $x, y > 0$,$$ g\left( \frac{x}{y} \right) = \left( \frac{x}{y} + 1 \right)^a - \left( \frac{x}{y} \right)^a \geqslant 1 \Longrightarrow (x + y)^a \geqslant x^a + y^a. $$ And the inequality is obviously true if either $x = 0$ or $y = 0$. $\square$

Now return to the proposition. First, the minimum is attainable since $\{\vec{x} \in \R_{\geqslant 0}^n \mid \|\vec{x}\|_a = t\}$ is tight in $\R^n$, i.e. closed and bounded.

On the one hand, taking $\vec{x} = (t, 0, \cdots, 0)$ shows that $\min\limits_{\substack{\vec{x} \in \R_{\geqslant 0}^n\\\|\vec{x}\|_a = t}} \|\vec{x}\|_1 \leqslant t$. On the other hand, the lemma implies that for $\vec{x} \in \R_{\geqslant 0}^n$ with $\|\vec{x}\|_a = t$,$$ \|\vec{x}\|_1^a = \left( \sum_{k = 1}^n x_k \right)^a \geqslant \left( \sum_{k = 1}^{n - 1} x_k \right)^a + x_n^a \geqslant \cdots \geqslant \sum_{k = 1}^n x_k^a = \|\vec{x}\|_a^a = t^a, $$ thus $\|\vec{x}\|_1 \geqslant t$ and $\min\limits_{\substack{\vec{x} \in \R_{\geqslant 0}^n\\\|\vec{x}\|_a = t}} \|\vec{x}\|_1 \geqslant t$. Therefore, $\min\limits_{\substack{\vec{x} \in \R_{\geqslant 0}^n\\\|\vec{x}\|_a = t}} \|\vec{x}\|_1 = t$. $\square$


As an example in which a non-linear function is the best lower bound, consider the inequality$$ x^2 + y^2 + 2 \geqslant f(x + y).\quad \forall (x, y) \in \R^2 $$ An obvious linear choice for $f$ is $f(t) = 2t$ since$$ (x^2 + y^2 + 2) - 2(x + y) = (x - 1)^2 + (y - 1)^2 \geqslant 0, $$ but the best bound is $f(t) = \dfrac{t^2}{2} + 2 \geqslant 2t$ because for any $t \in \R$,$$ (x^2 + y^2 + 2)\bigr|_{x + y = t} = x^2 + (t - x)^2 + 2 = 2\left( x - \frac{t}{2} \right)^2 + \frac{t^2}{2} + 2 \geqslant \frac{t^2}{2} + 2, $$ and the equality is attained when $x = y = \dfrac{t}{2}$.

$\endgroup$
3
  • 1
    $\begingroup$ If I understood correctly, this proves that best function (either linear or non-linear) $f$ of $QM$ such that $AM \le f(QM)$ is $f(t)=\frac{t}{\sqrt{n}}$. $\endgroup$ – ands May 13 '20 at 11:49
  • $\begingroup$ Yes, of course. $\endgroup$ – ands May 13 '20 at 12:11
  • $\begingroup$ @ands I've updated the answer with an example for demonstration. $\endgroup$ – Saad May 13 '20 at 13:46
0
$\begingroup$

Let we need to find a maximal $C(n)$, for which the inequality $$\left(\sum_{k=1}^nx_k\right)^2\geq C\sum_{k=1}^nx_k^2$$ is true for any non-negatives $x_k$.

Let $x_2=x_3=...=x_n=0$.

Thus, $C\leq1,$ which says $C=1$ is a best bound.

There is the following inequality:

For any $x_i\geq0$, $n\geq2$ prove that: $$\sum_{i=1}^nx_i\leq\sqrt{\frac{\sum\limits_{1\leq i<j\leq n}x_ix_j}{\binom{n}{2}}}+(n-1)\sqrt{\frac{\sum\limits_{i=1}^nx_i^2}{n}}.$$

It's stronger because $$\frac{\sum\limits_{i=1}^nx_i}{n}\geq\sqrt{\frac{\sum\limits_{1\leq i<j\leq n}x_ix_j}{\binom{n}{2}}}.$$ For specific values of $n$ we can get much more stronger inequalities.

$\endgroup$
3
  • 1
    $\begingroup$ This is interesting inequality, but the problem is that I dont know $\sqrt{\frac{\sum\limits_{1\leq i<j\leq n}x_ix_j}{\binom{n}{2}}}$. $\endgroup$ – ands May 12 '20 at 18:53
  • $\begingroup$ @ands We know: $\sum\limits_{1\leq i<j\leq n}x_ix_j=\frac{1}{2}\left(\left(\sum\limits_{i=1}^nx_i\right)^2-\sum\limits_{i=1}^nx_i^2\right)=\frac{1}{2}(n^2A^2-nQ^2).$ $\endgroup$ – Michael Rozenberg May 12 '20 at 19:04
  • 2
    $\begingroup$ Yes, but I only now quadratic mean ($Q$). I don't know arithmetic mean ($A$), that's why I am asking for lower bound for arithmetic mean. $\endgroup$ – ands May 12 '20 at 22:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.