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Show that $\max{\{|a|+|b|,|c|+|d|\}} \leq \max{\{|a|,|c|\}}+\max{\{|b|,|d|\}}.$

I wanted to show that $d(p,q)=\max{\{|x_1-x_2|,|y_1-y_2|\}}$ where $p=(x_1,y_1),q=(x_2,y_2)$ is a metric on $\mathbb{R^2}.$ In proving the triangle inequality let $p=(x_1,y_1),q=(x_2,y_2),r=(x_3,y_3).$ Then \begin{align*} d(p,r)&=\max{\{|x_1-x_3|,|y_1-y_3|\}}\\ &\le \max{\{|x_1-x_2|+|x_2-x_3|,|y_1-y_2|+|y_2-y_3|\}}\\ &\le \max{\{|x_1-x_2|,|y_1-y_2|\}}+\max{\{|x_2-x_3|,|y_2-y_3|\}}=d(p,q)+d(q,r) \end{align*} I don't understand how this inequality is true.

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Hint: Try to write

$$\max(f,g)=\frac{|f-g|+f+g}{2}.$$

Maybe can help.

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$|a| \le \max(|a|,|b|)$ and $|c| \le \max(|c|,|d|)$ so $|a| + |b| \le \max(|a|,|b|) + \max(|c|,|d|)$.

Likewise $|b| \le \max(|a|,|b|)$ and $|d| \le \max(|c|,|d|)$ so $|b| + |d| \le \max(|a|,|b|) + \max(|c|,|d|)$.

So both $|a| + |c|$ and $|b| + |d|$ $\le \max(|a|,|b|) + \max(|c|,|d|)$.

So $\max(|a| + |c|,|b| + |d|)\le \max(|a|,|b|) + \max(|c|,|d|)$.

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It all boils down to the fact that for any real number $a$, $|a| \le \max\{|a|, |b|\}$ for any real number $b$. Let me do the first inequality. Suppose $\max\{|x_1-x_3|,|y_1-y_3|\}=|x_1-x_3|$, triangle inequality yields \begin{align*} \max\{|x_1-x_3|,|y_1-y_3|\}=|x_1-x_3| & \le |x_1-x_2| + |x_2-x_3|\\ & \le \max\{|x_1-x_2| + |x_2-x_3|, |y_1-y_2|+|y_2-y_3|\}. \end{align*} If on the other hand, $\max\{|x_1-x_3|,|y_1-y_3|\}=|y_1-y_3|$, a similar argument yields \begin{align*} \max\{|x_1-x_3|,|y_1-y_3|\}=|y_1-y_3| & \le |y_1-y_2| + |y_2-y_3|\\ & \le \max\{|y_1-y_2| + |y_2-y_3|, |x_1-x_2|+|x_2-x_3|\}. \end{align*} Thus, we see that in both cases, we have $$ \max\{|x_1-x_3|,|y_1-y_3|\}\le \max\||x_1-x_2|+|x_2-x_3|,|y_1-y_2|+|y_2-y_3|\}. $$ The second inequality follows by a similar argument.

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