# Lower bound for the arithmetic mean based on quadratic mean

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 $$QM$$?

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 $$0$$.

$$\def\vec{\boldsymbol}\def\R{\mathbb{R}}$$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 $$x = y = \dfrac{t}{2}$$.

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

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

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

• This is interesting inequality, but the problem is that I dont know $\sqrt{\frac{\sum\limits_{1\leq i<j\leq n}x_ix_j}{\binom{n}{2}}}$. – ands May 12 '20 at 18:53
• @ands We know: $\sum\limits_{1\leq i<j\leq n}x_ix_j=\frac{1}{2}\left(\left(\sum\limits_{i=1}^nx_i\right)^2-\sum\limits_{i=1}^nx_i^2\right)=\frac{1}{2}(n^2A^2-nQ^2).$ – Michael Rozenberg May 12 '20 at 19:04
• 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. – ands May 12 '20 at 22:48