Calculus: Proofs Help with homework: If a < 0, b < 0, then √ab ≤ −(a+b)/2

[mathkid11]

Calculus: Proofs Help with homework: If a < 0, b < 0, then √ab ≤ −(a+b)/2

[h=1]How do I prove this statement using properties of numbers: If a < 0, b < 0, then √ab ≤ −(a+b)/2?[/h]

 

[Deleted member 4993]

How do I prove this statement using properties of numbers: If a < 0, b < 0, then √(ab) ≤ −(a+b)/2?

If you do not put those (), we would have:

√ab = b√a

What are your thoughts?

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[stapel]

How do I prove this statement using properties of numbers: If a < 0, b < 0, then √ab ≤ −(a+b)/2?

What "properties of numbers" have they given you to use? How far have you gotten in applying this information?

For instance, you noted that, if you square both sides of the proposed inequality (not a valid "proof" step, but often helpful when doodling and trying stuff), you got "ab" from the left-hand side (LHS) and [(a^2 + 2ab + b)^2]/4 on the right-hand side (RHS). If you multiply both by 4, you get 4ab and a^2 + 2ab + b^2, respectively. If you subtract 4ab from each expression, you get 0 and a^2 - 2ab + b^2, which equals (a - b)^2, which is always non-negative. In other words, you end up with something that you could state as a true inequality:

. . . . .\(\displaystyle 0\, \leq \, (a\, -\, b)^2\)

Once you discovered this, how did you use it in attempting to develop a proper proof, starting with the "if" statement?

Please be complete. Thank you!