Let's start by expanding the left side of the inequality:
(abs(A+B)(A-B)) = (abs(A) + abs(B))(A-B)
Using the distributive property, we can expand this expression:
= abs(A)(A-B) + abs(B)(A-B)
Using the distributive property again, we have:
= A*abs(A) - B*abs(A) + A*abs(B) - B*abs(B)
Now, let's expand the right side of the inequality:
(abs(A))^2 + (abs(B))^2 = A*abs(A) + B*abs(B)
Now, let's compare the two sides of the inequality:
A*abs(A) - B*abs(A) + A*abs(B) - B*abs(B) ≤ A*abs(A) + B*abs(B)
Rearranging the terms, we have:
-A*abs(A) - B*abs(B) ≤ 0
Multiplying both sides by -1, we get:
A*abs(A) + B*abs(B) ≥ 0
Since this inequality holds true for all values of A and B, we can conclude that:
abs(A+B)(A-B) ≤ (abs(A))^2 + (abs(B))^2
Therefore, abs(A+B)(A-B) is less than or equal to abs(A)^2 + abs(B)^2, which proves the inequality.
Show that abs(A+B)(A-B)<=abs(A)sqr+abs(B)sqr
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