Could anyone advise me how to find $\text{sup} \{|x_1 -y_1| + |x_2 - y_2| + |x_3 -y_3| : \ x_1^2 + x_2^2 + x_3^2 \leq 1, y_1^2 + y_2^2 + y_3^2 \leq 1 \} \ ?$ Thank you.
Here is my attempt:
$(|x_1 -y_1| + |x_2 - y_2| + |x_3 -y_3|)^2$ $= x_1^2+x_2^2 +x_3^2 + y_1^2+y_2^2 + y_3^2 + 2[(x_1-y_1)(x_2-y_2) +(x_1-y_1)(x_3-y_3) + (x_2-y_2)(x_3-y_3)] \leq 2 \ + \ ... \ ?$
Observe that $(x-y)^2 = x^2+y^2-2xy \leq 2(x^2+y^2)$ with equality if and only if $x=-y$.
Using the Jensen ineqaulity and above, we get that \begin{align*} (|x_1-y_1|+|x_2-y_2| + |x_3-y_3|)^2 &\leq 3\left((x_1-y_1)^2+(x_2-y_2)^2 + (x_3-y_3)^2\right) \\ &\leq 6\left(x_1^2+x_2^2+x_3^2 + y_1^2+y_2^2+y_3^2\right)\\ &\leq 12. \end{align*}
Backtracking, we see that we get equality exactly when $x_i=-y_i$, the differences are equal and $x_1^2+x_2^2+x_3^2=1$ and $y_1^2+y_2^2+y_3^2=1$. This is possible by taking $x_i=1/\sqrt{3}$ and $y_i=-x_i$.
Hence the diameter is $2\sqrt{3}$.