# Is the product of numbers always less than or equal to the sum of squares?

I would like to know whether the inequality $x_1 x_2 ... x_n \leq x_1^2 + x_2^2 + ... + x_n^2$, i.e. $\prod_{i=1}^n x_i \leq \sum_{i=1}^n x_i^2$ holds for all $x_1, x_2, ..., x_n \in \mathbb{R}$ where $n \geq 2$.

Clearly it doesn't hold for $n=1$; we have the inequality $x \leq x^2, \, x \in \mathbb{R}$, so $x=1/2$ is a counterexample for instance. It does however hold for $n=2$. We have the inequality $xy \leq x^2+y^2, \, x, y \in \mathbb{R}$. It holds trivially if $x$ or $y$ is zero. If $x$ and $y$ differ in sign, it also holds. In fact we only have to check the case $x,y>0$ since both sides of the inequality are unchanged if we send $(x, y) \mapsto (-x, -y)$. So let $x, y>0$. Then $$xy \leq x^2+y^2$$ $$\Longleftrightarrow \ -xy \leq x^2 - 2xy + y^2 = (x-y)^2$$ which is true since $-xy < 0 \leq (x-y)^2$.

For $n=3$, I don't know whether or not the inequality holds. I looked for a counterexample for a while by doing some 3D plots, but GeoGebra wasn't playing along. We have the inequality $xyz \leq x^2+y^2+z^2, \, x,y,z \in \mathbb{R}$. Again, it holds trivially if any of $x, y, z$ are zero. We have the following possibilities for the signs of $x, y, z$: $(+, +, +), (+, +, -), (+, -, +), (+, -, -), (-, +, +), (-, +, -), (-, -, +), (-, -, -)$.

If there is an odd number of $-$ signs then the inequality holds, which leaves us with $(+, +, +), (+, -, -), (-, +, -), (-, -, +)$, however both sides of the inequality remain unchanged if we send $(y, z) \mapsto (-y, -z)$, $(x, z) \mapsto (-x, -z)$, or $(x, y) \mapsto (-x, -y)$. Hence we only need to check for $x, y, z > 0$.

If it holds for $n=3$, my intuition tells me it probably holds for all $n \geq 2$, and that there is a fairly simple way of showing this by induction.

Thanks.

• $4\cdot 4\cdot 4>4^2+4^2+4^2$ – Wojowu Feb 13 '18 at 19:29
• For others, Let $x_i=n$ for $n>3$… – Macavity Feb 13 '18 at 19:35
• COnsider the special case $x_1 = x_2 = \dots = x_n$. Then you should have $$x^n \le nx^2$$ and this is clearly false as $x \to + \infty$. – Crostul Feb 13 '18 at 19:39
• For large $n$, $n!$ is much larger than $1^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$. – Daniel Schepler Feb 13 '18 at 19:39
• You do have in general the similar result (for positive numbers) that geometric mean $\le$ root mean square: $\sqrt[n]{x_1 \cdots x_n} \le \sqrt{ \frac{x_1^2 + \cdots + x_n^2}{n} }$. – Daniel Schepler Feb 13 '18 at 19:41

$\require{cancel}$ $\displaystyle$

You cannot conclude it for $n \geq 2$, as simply the power of left side is $n$ and the right side is $2$, thus:

$O(n) \cancel{\leq} O(2)$ for $n \geq 2$

In a simple example, if $x_1=x_2=...=x_n=x$, then $x^n \cancel{\leq} nx$.

What you can say instead of that, is the general inequity of this:

$\prod_{i=1}^n x_i \leq \sum_{i=1}^n x_i^n$

In your case (i.e. when $n=2$) it will become the specific inequity you mentioned first:

$\prod_{i=1}^2 x_i \leq \sum_{i=1}^2 x_i^2 \Longleftrightarrow xy \leq x^2+y^2$

Note that

$$3\times 4\times 5= 60$$ $$3^2+4^2+5^2 =50$$ Thus $$x_1 x_2 ... x_n \leq x_1^2 + x_2^2 + ... + x_n^2$$ is not true for all real numbers.

wolog we may assume $x_i > 0$. (If an odd number are negative then the product is zero or less while the sum of squares is positive. If an even number are negative the product and the sum of squares will be the same as though we had only used absolute values. And if any are zero the product will be zero which is less or equal to the sum of the squares.)

If $x_1x_2 \le x_1^2 + x^2 \iff 1 \le \frac {x_1}{x_2} + \frac {x_2}{x_1}$ which be the AM-GM is always true. ($r + \frac 1r \ge 2$ for all $r \in \mathbb R^+$.)

But we have no such luck with $\prod x_i \le \sum x_i^2$. We know by $AM-GM$ that $\prod x_i \le \sum \frac {x_i^n}n$ (which equality holding if and only if the $x_i$ are equal). But it's very possible for $\sum x_i^2 < \prod x_i \le \sum \frac {x_i^n}n$.

For example if $x_i=x$ are equal and $x > \sqrt[n-2]{n}$. (Then $\prod x_i = x^n = (x^{n-2})*x^2 > nx^2 = x_1^2+ ... +x_n^2$.)

Exceptions are exceedingly easy to come by. If the $x_i$ are relatively close together than $\sum x_i^2 \ge \sqrt[n]{\prod x_i^2}$ but not necessarily by a huge amount. So on the whole if $x_i$ are significantly larger than $\sqrt[n-2]{n}$ it'll do.

Example: if $x,y,z$ are larger than $3$ so say $6,4,9$ we have $6*4*9 = 216$ while $36 + 16 + 81 = 133$ is a lot smaller. $\sqrt[3]5 \approx 1.7$ so $2,3,2,4,3$ is such $\prod x_i = 144$ whereas $\sum x_i^2 = 52$. We can even have an $x_i < 1.7$ if the others are large enough $1, 2,3,3,4$ is such $\prod x_i = 72$ and $\sum x_i = 39$.

For $n=2$ this easily arises from the fact that $$(x_1-x_2)^2\ge 0$$ However, for $n\geq 3$ it is extremely false. This is actually easy to see. Set $x_i=x$ for all $i$. Then you're trying to prove that $$x^n\leq nx^2$$ For large $x$ this inequality doesn't hold.