I found an inequality as following: Let $x, y, z$ be three complex numbers then:

\begin{equation*} \frac{1}{2}(|y+z-x|+|x+z-y| + |y+x-z|) \le |x| + |y|+|z|+\frac{1}{2}|x+y+z| \end{equation*} (1)

The inequality holds with equality if and only if $x+y+z=0$

Note that: I have a proof of the inequality (1).

My question: I am looking for a proof of conjecture as following:

 

Let $x, y, z$ in an inner product space $V$ then

 

\begin{equation*}\frac{1}{2}(\|y+z-x\|+\|x+z-y\| + \|y+x-z\|) \le \|x\| + \|y\|+\|z\|+\frac{1}{2}\|x+y+z\|\end{equation*}

 

where the norm ||z|| denotes the norm induced by the inner product

See also

• Hlawka's inequality
• Absolute value inequality for complex numbers

I found an inequality as following: Let $x, y, z$ be three complex numbers then:

\begin{equation*} \frac{1}{2}(|y+z-x|+|x+z-y| + |y+x-z|) \le |x| + |y|+|z|+\frac{1}{2}|x+y+z| \end{equation*} (1)

The inequality holds with equality if and only if $x+y+z=0$

Note that: I have a proof of the inequality (1).

My question: I am looking for a proof of conjecture as following:

Let $x, y, z$ in an inner product space $V$ then

\begin{equation*}\frac{1}{2}(\|y+z-x\|+\|x+z-y\| + \|y+x-z\|) \le \|x\| + \|y\|+\|z\|+\frac{1}{2}\|x+y+z\|\end{equation*}

where the norm ||z|| denotes the norm induced by the inner product

See also

• Hlawka's inequality
• Absolute value inequality for complex numbers

I found an inequality as following: Let $x, y, z$ be three complex numbers then:

\begin{equation*} \frac{1}{2}(|y+z-x|+|x+z-y| + |y+x-z|) \le |x| + |y|+|z|+\frac{1}{2}|x+y+z| \end{equation*} (1)

The inequality holds with equality if and only if $x+y+z=0$

Note that: I have a proof of the inequality (1).

My question: I am looking for a proof of conjecture as following:

Let $x, y, z$ in an inner product space $V$ then

\begin{equation*}\frac{1}{2}(\|y+z-x\|+\|x+z-y\| + \|y+x-z\|) \le \|x\| + \|y\|+\|z\|+\frac{1}{2}\|x+y+z\|\end{equation*}

where the norm ||z|| denotes the norm induced by the inner product

See also

• Hlawka's inequality
• Absolute value inequality for complex numbers

I found an inequality as following: Let $x, y, z$ be three complex numbers then:

\begin{equation*} \frac{1}{2}(|y+z-x|+|x+z-y| + |y+x-z|) \le |x| + |y|+|z|+\frac{1}{2}|x+y+z| \end{equation*} (1)

The inequality holds with equality if and only if $x+y+z=0$

Note that: I have a proof of the inequality (1).

My question: I am looking for a proof of conjecture as following:

Let $x, y, z$ in an inner product space $V$ then

\begin{equation*}\frac{1}{2}(\|y+z-x\|+\|x+z-y\| + \|y+x-z\|) \le \|x\| + \|y\|+\|z\|+\frac{1}{2}\|x+y+z\|\end{equation*}

where the norm ||z|| denotes the norm induced by the inner product

See also

• Hlawka's inequality
• Absolute value inequality for complex numbers