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

Question

Calculus: Proofs

Answer 1

Given: a<0; b<0

Prove : -(a+b)/2 >= sqrt(ab)

a<0 means a^2 >0
b<0 means b^2 >0

Then ab>0 and a^2 + b^2 > 0
let t = sqrt(ab)

a+b<0 means -(a+b)>0
so -(a+b)/2 > 0

The proof will be completed in two parts,
by contradiction, first with t=sqrt(ab)>0,
then with t=sqrt(ab)<0

Suppose 0 <-(a+b)/2 < t =sqrt(ab).

0 < -(a+b) < 2*t = 2*sqrt(ab) <--- multiplies everything by 2

0 > a+b > -2*t = -2*sqrt(ab) <-- multiplies everything by -1

0 > (a+b)^2 > 4ab <-- squaring everything is allowed

0 > a^2 + 2ab + b^2 > 4ab <-- FOIL ; squaring binomial

-2ab > a^2 + b^2 > 2ab <--- subtracts 2ab from everything

This is a contradiction because ab>0 and a^2 +b^2 >0 while -2ab <0.
This completes the proof for sqrt(ab)>0


Suppose 0 > t=sqrt(ab) > -(a+b)/2.

0 > 2*t = 2*sqrt(ab) > -(a+b) <--- multiplies everything by 2

0 < -2*sqrt(ab) < (a+b) <--- multiplies everything by -1

But 0 > a+b, because a<0 and b<0.
So 0 < -2*sqrt(ab) < a+b < 0.

Moreover -2*sqrt(ab) > 0 since sqrt(ab) <0 by supposition.

This is a big-time, train wreck contradiction: 0 < positive < negative < 0. 

So either t = sqrt(ab)<0 is false or sqrt(ab)> -(a+b)/2

If t = sqrt(ab)<0 is false, then t=sqrt(ab)>=0 in which
case the proof is complete as proven above.

Otherwise the latter statement is false which means
sqrt(ab) <= -(a+b)/2 which completes the proof.

Answer 2

Consider (a+b)²; this will be the square of a negative number, and thus positive.

(a+b)² = a² + 2ab + b² [algebraic identity in which all right-side terms are positive]

 

If we eliminate the first & last terms of the above right side, then certainly (a+b)² ≥ 2ab.

Taking principal square roots, we have |a+b| ≥ √2•√(ab).

Dividing, we get |a+b| / √2 ≥ √(ab).

Replacing |a+b| by -(a+b) because both a & b are > 0, we have -(a+b) / √2  ≥ √(ab).

 

Now, note that I proved an assertion different from what you wanted--yours has the INTEGER 2 in the denominator.

 

I welcome feedback from all who are interested in this.