Andy C. answered • 09/17/17

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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.

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.

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.

Kenneth S.

09/17/17