I checked these

What is the difference between square of sum and sum of square?

Prove $(|x| + |y|)^p \le |x|^p + |y|^p$ for $x,y \in \mathbb R$ and $p \in (0,1]$.

It is easy to see $p$-th power ($p\ge 1$) version, i.e., $|x|^p+|y|^p\le (|x|+|y|)^p$ ($p\ge 1$), holds as well using the argument by Quang Hoang in Prove $(|x| + |y|)^p \le |x|^p + |y|^p$ for $x,y \in \mathbb R$ and $p \in (0,1]$. Is there a name for this inequality so that I can just quote? (It might be an elementary result but people around me bother to put "from Cauchy--Schwarz,..." when it is clearly Cauchy--Schwarz, so.)

  • 1
    $\begingroup$ You can say that the function $x\mapsto x^p$ on $[0,\infty)$ is subadditive if $p\in(0,1]$ and superadditive if $p\geq1$. $\endgroup$ – triple_sec Aug 3 '15 at 1:39
  • 3
    $\begingroup$ I've sometimes heard, "by comparison of $\ell_p$ norms". (This is the two-dimensional case of $\ell_p$ vs $\ell_1$.) $\endgroup$ – user21467 Aug 3 '15 at 1:43
  • $\begingroup$ In probability it's called $c_r$ inequality when you put expectations on both sides... $\endgroup$ – d.k.o. Aug 3 '15 at 3:20
  • $\begingroup$ @d.k.o. As in www2.cirano.qc.ca/~dufourj/Web_Site/ResE/…? I am not sure if I am looking at a right source but the constant and the direction look a bit different though. $\endgroup$ – shall.i.am Aug 3 '15 at 10:36

I do not believe there is a name for your specific inequality (which I rewrite as following): $$ |x|^p+|y|^p\le \big(|x|+|y|\big)^p, \ p \ge1 \iff \boxed{\ \ \left(|x|^p +|y|^p\right)^{\frac{1}{p}} \le |x|+|y|, \quad p \ge 1 \ \ } $$ However, it can be viewed as a special case of multiple more general statements, such as Jensen, AMGM, Hölder, and probably many other inequalities after appropriate substitution and/or change of variables. The closes call would probably be the generalized mean inequality: $$ M_j\left( x_1, \dots, x_n \right) \leq M_i\left( x_1, \dots, x_n \right) \quad \text{ whenever } \quad j<i. \label{*} \tag{*} $$

Here $M_k \left( x_1, \dots, x_n \right)$ is so-called power mean, which is defined as $$ M_k(x_1,\dots,x_n) = \left( \frac{1}{n} \sum_{i=1}^n x_i^k \right)^{\frac{1}{k}}. $$

! In particular, assuming $n=2$, $j = 1$, and $i = p$, and denoting $\left(x_1, \dots, x_n \right) := \left(\,\chi, \gamma \right)$, we get $$ \begin{aligned} M_1\big(\left|\,\chi\right|, \left|\gamma\right|\big) & = \dfrac{1 }{2}\big(\left|\,\chi\right| + \left|\gamma\right|\big) = \dfrac{\left|\,\chi\right| }{2} + \dfrac{\left|\gamma\right| }{2}, \\ M_p\big(\left|\,\chi\right|,\left|\gamma\right|\big) & = \left( \dfrac{1}{2} \left(\left|\,\chi\right|^p+\left|\gamma\right|^p\right)\right)^{\frac{1}{p}} = \Bigg( \left(\frac{\left|\,\chi\right|}{2}\right)^p + \left(\frac{\left|\gamma\right|}{2}\right)^p \Bigg)^{\frac{1}{p}}.\\ \end{aligned} $$ By $\eqref{*}$, we have $$ \dfrac{\left|\,\chi\right| }{2} + \dfrac{\left|\gamma\right| }{2} \le \left( \bigg(\frac{\left|\,\chi\right|}{2^{\frac{1}{p}}}\bigg)^p + \bigg(\frac{\left|\gamma\right|}{2^{\frac{1}{p}}}\bigg)^p \right)^{\frac{1}{p}} . \label{**} \tag{**} $$ Denoting $ x := 2^{-\frac{1}{p}}\chi, \ \ y := 2^{-\frac{1}{p}}\gamma$ and raising both sides of $\eqref{**}$ to the power $p$, we get
$$ \left|x\right|^{p}+\left|y\right|^{p} \le \big(\left|x\right|+\left|y\right|\big)^{p}. $$

To summarize, I believe that (strictly speaking) there is probably no name for your inequality. However, that the generalized mean is as close as you can get to your inequality, although some formula conversion is still required.


I think this should be called an special case of Minkowski's Inequality.

1 - For finite sequences, the Minkowski Inequality states that

$$ \left(\sum_{k=1}^n |x_k + y_k |^p \right) ^{1/p} \le \left(\sum_{k=1}^n |x_k|^p \right)^{1/p} + \left(\sum_{k=1}^n |y_k|^p\right)^{1/p} $$

Your inequality then follows if $x_1 = x, x_i = 0, \; i \ge 2$, and $y_2 = y, y_i = 0, i \ne 2$.

2 - More Generally, the case above can be easily extended to infinite sequences. The most general result is that, if $f,g : \Omega \to \Bbb{R}$ are measurable functions in a measure space $(\Omega, \Sigma, \mu)$, then the inequality

$$ \| f+g \|_{L^p (d\mu)} \le \|f\|_{L^p(d\mu)} + \|g\|_{L^p(d\mu)} $$

Where $ \|f\|_{L^p(d\mu)}^p = \int_ {\Omega} |f(x)|^p d \mu (x) $

3 - There was a little lie at topic 2, because there is yet another generalization of this result, called Minkowski's Inequality for Integrals. In the link above, you might check a proof of this result, based on these previously presented ones.

  • 1
    $\begingroup$ Ah. It does follow from Minkowski. You meant $y_i=0$ $i\neq 2$, right? Namely, $\boldsymbol{x}:=(x_1,0,\dotsc)$ and $\boldsymbol{y}:=(0,y_2,0,\dotsc)$. From Minkowski, $(|x_1|^p+|y_2|^p)^{\frac1p}\le|x_1|^{\frac1pp}+|y_2|^{\frac1pp}$ and thus $(|x_1|^p+|y_2|^p)\le(|x_1|+|y_2|)^p$. I did not realise this because if anything Minkowski looks like the opposite direction. I find your answer really interesting but just saying "this follows from Minkowski" sounds a bit misleading as it looks opposite, so, sorry! But really appreciate it! $\endgroup$ – shall.i.am Aug 3 '15 at 10:33
  • $\begingroup$ Yeah, it really looks opposite, right? I was thinking about this inequality about two weeks ago, when I realized that it was just Minkowski's Inequality. Well, just wanted to help here :) Be free to ask more questions about this answer if you like :D $\endgroup$ – João Ramos Aug 3 '15 at 12:38

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.