Proving σmax(A-B) ≤ σmax(A) - σmin(B)

[hayu601]

Is it true if I state:

σ_max(A-B) <= σ_max(A) - σ_min(B) ?

I verify numerically that it is correct but how to prove it?

 

[HallsofIvy]

If $a\in A$ then $a\le max(A)$
If $b\in B$ then $b\ge min(a)$ so $-b\ge -min(B)$

Adding, $a-b\le max(A)- min(B)$ for all a in A, b in B/