# $\forall\ x,y,z\in \mathbb{R}$ Show that: $|x+y|+|y+z|+|x+z|\leq |x+y+z|+|x|+|y|+|z|$

$\forall\ x,y,z\in \mathbb{R}$ Show that: $$|x+y|+|y+z|+|x+z|\leq |x+y+z|+|x|+|y|+|z|$$

i tired,

i notice that $x,y,z$ plays a symmetrical role in the inequality

notice also that

\begin{align*} |x+y|+|y+z|+|x+z|\leq |x+y+z|+|x|+|y|+|z| & \Longleftrightarrow \\ (|x+y|-|x|)+(|y+z|-|y|)+(|x+z|-|z|) \leq |x+y+z| \end{align*} note that $\forall a,b\in \mathbb{R}\quad |a|-|b|\leq |a+b|$ then $$(|x+y|-|x|)+(|y+z|-|y|)+(|x+z|-|z|) \leq |x|+|y|+|z|$$ i'm stuck here

any help would be appreciated!

Let $a=x+y$, $b=x+z$ and $c=y+z$. Then, the inequality to be shown can be rewritten as

$$2(|a|+|b|+|c|) \leq |a+b+c|+|a+b-c|+|a-b+c|+|-a+b+c| \tag{1}$$

Let us put $f(a,b,c)=|a+b+c|+|a+b-c|+|a-b+c|+|-a+b+c|$. It is easy to see that $f$ is even in each variable, so that $f(a,b,c)=f(|a|,|b|,|c|)$. We may therefore assume that $a,b,c$ are all nonnegative, so that it suffices to show that

$$2(a+b+c) \leq |a+b+c|+|a+b-c|+|a-b+c|+|-a+b+c|\tag{2}$$

But (2) follows from the triangle inequality, since

$$2(a+b+c)=(a+b+c)+(a+b-c)+(a-b+c)+(-a+b+c) \tag{3}$$

• I wonder if this holds in $\mathbb C$, I needed we are in $\mathbb R$ in my solution too, but I won't post it, since this is much nicer. – user2345215 Nov 26 '14 at 18:34
• @user2345215 It holds for any inner product space as well, known as the Hlawka's Inequality – sciona Nov 26 '14 at 18:38
• @sciona Neat! Thanks. – user2345215 Nov 26 '14 at 18:39
• @Educ yes, right! – sciona Nov 27 '14 at 6:12
• one of professor that i know told me that i don't have right to say if $f$ is even in each variable, so that $f(a,b,c)=f(|a|,|b|,|c|)$. is there any theorem or way to convince him that we can right $f(a,b,c)=f(|a|,|b|,|c|)$ he told me we have only even for every variable – Educ Nov 27 '14 at 17:41

We need to prove that $$\left(|x+y+z|+|x|+|y|+|z|\right)^2\geq\left(|x+y|+|x+z|+|y+z|\right)^2$$ or $$\sum_{cyc}\left(|x(x+y+z)|+|yz|\right)\geq\sum_{cyc}|(x+y)(x+z)|,$$ which is just a triangle inequality: $$|x(x+y+z)|+|yz|\geq|x(x+y+z)+yz|=|(x+y)(x+z)|.$$

Also we can use Popoviciu.

For all convex function $f$ we have $$f(x)+f(y)+f(z)+3f\left(\frac{x+y+z}{3}\right)\geq2\left(f\left(\frac{x+y}{2}\right)+f\left(\frac{x+z}{2}\right)+f\left(\frac{y+z}{2}\right)\right).$$ Since $f(x)=|x|$ is a convex function, we obtain $$|x|+|y|+|z|+3\left|\frac{x+y+z}{3}\right|\geq2\left(\left|\frac{x+y}{2}\right|+\left|\frac{x+z}{2}\right|+\left|\frac{y+z}{2}\right|\right)$$ or $$|x+y+z|+|x|+|y|+|z|\geq|x+y|+|x+z|+|y+z|$$ and we are done!

Prove that $\|a\|+\|b\| + \|c\| + \|a+b+c\| \geq \|a+b\| + \|b+c\| + \|c +a\|$ in the plane.

Note that indentity： \begin{align*} &(|a|+|b|+|c|-|b+c|-|a+c|-|a+b|+|a+b+c|)(|a|+|b|+|c|+|a+b+c|)\\ &=(|b|+|c|-|b+c|)(|a|-|b+c|+|a+b+c|)+(|c|+|a|-|c+a|)(|b|-|c+a|+|a+b+c|)+(|a|+|b|-|a+b|)(|c|-|a+b|+|a+b+c|) \end{align*}

• – Educ Dec 9 '14 at 8:25