Show that abs(A+B)(A-B)<=abs(A)sqr+abs(B)sqr

Answer 1

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.