Give the proof of the inequality $0.5(|A+B|+|A-B|)= \max (|A|,|B|)$ $$0.5(|A+B|+|A-B|)= \max (|A|,|B|)$$
where $A$ and $B$ are any values (real number)
I have tried to solve this using triangle inequality but got $0.5(|A+B|+|A-B|)> |A|$
How to prove the given inequality??
 A: WLOG, $A\ge B\ge0$ because swapping $A,B$ and/or changing one or two signs has no effect.
Then the identity reduces to
$$\frac{A+B+A-B}2=A,$$ which is true.
A: I will assume here that you meant $0.5(|A+B|+|A-B|) \le \max \{|A|,|B|\}$ rather than $0.5(|A+B|+|A-B|) = \max \{|A|,|B|\}.$
As often happens with absolute values, it is useful to consider two cases:

*

*$A,B$ have the same sign (i.e. both are positive or both negative).

*$A,B$ have different signs.

No generality is lost by assuming $|A|\ge|B|,$ i.e. whichever one of $A,B$ has the larger absolute value will be called $A$ and the other $B.$  Then $\max\{|A|,|B|\} = |A|.$
If $A,B$ are both positive you have
\begin{align}
& 0.5(|A+B|+|A-B|) \\[6pt]
= {} & 0.5(A+B) + 0.5|A-B| \\[6pt]
\le {} & 0.5(A+B) \le 0.5(A+A) = A = \max\{|A|,|B|\}.
\end{align}
If both are negative then a similar argument works.
Now suppose one is positive and one negative. Then $|A-B| = |A|+|B|,$ and so
\begin{align}
& 0.5(|A+B|+|A-B|) \\[6pt]
= & 0.5(|A|+|B|) + 0.5|A+B| \\[6pt]
\le {} & 0.5(|A|+|A|) = |A| = \max\{|A|,|B|\}.
\end{align}
