If $a$ and $b$ are positive real numbers, then $a + b \geq 2 \sqrt{ab}$.

Question

If $a$ and $b$ are positive real numbers, then $a + b \geq 2 \sqrt{ab}$.

I know how to do the direct proof, but in this case, I want to try proving it by contradiction. I have tried manipulating the inequality $a + b < 2 \sqrt{ab}$ after making the assumption that $a,b >0$ to get a contadiction $a,b \leq 0$.

$\begin{align} a + b &< 2 \sqrt{ab} \\a^2-2ab+b^2 &< 0 \\(a-b)^2 &< 0\end{align}$

How do I show that $a,b \leq 0$?

Answer 1

You are dealing with real numbers... It is impossible that $(a-b)^2<0$ because squaring a quantity always results in another quantity that is greater than or equal to zero.

EDIT: You seem to believe that a very specific contradiction is needed to complete this proof. Namely, that $a,b \leq 0$ has to be reached before you can stop. That isn't how proof by contradiction works. As long as you get any form of contradiction, you can end the proof. The fact that you got $(a-b)^2<0$ is perfect, because it violates one of your assumptions. You assumed $a,b$ are real numbers in addition to assuming that $a,b >0$. The only way you can square $a-b$ and end up negative is if $a-b$ is a complex number. That violates the assumption that $a,b$ are real, hence you have a contradiction.

Answer 2

1. assume that a,b<0 That would mean : (a+b)/2 = (-a+-b)/2 =(-a-b)/2 which is negative That would also mean: (ab)^1/2= (-a x -b)^1/2 = (ab)^1/2 which is positive hence,

   (-a-b)/2 < (ab)^1/2

2. now lets convert the above inequality so as to convert a,b<0 into a,b>0.

   To do that, we multiply both sides by -1 like so:

   -1{(-a-b)/2}< -1{(ab)^1/2} and that gives us

   (a+b)/2 > -(ab)^1/2 where a,b are positive integers .
   
   now we all know that if a,b>0, there would not be any negative sign 
   in of the the square root of (ab). so we put a modulus sign in front 
   of the negative suare root of (ab) like so:
   
   (a+b)/2 > |-(ab)^1/2| and   sice we know that modulus of any negative
   number is positive , we can remove both the modulus sign and the 
   negative sign which give us the following inequality:
   
    (a+b)/2 > (ab)^1/2. 
   

   we have proved that (a+b)/2 > (ab)^1/2

3. now we are left to introduce the equal sign in the inequality. We are going to do this by assuming that a and b are positve and equal as well.

   so lets say that a=b and replace a by b in the equations like so: (a+b)/2 and (ab)^1/2

    since a=b
    (b+b)/2 = 2b/2
            =b
   
     also ,
    (b x b)^1/2 = b
   

   hence we also proved that (a+b)/2 =(ab)^1/2.

   so in general , we can say that (a+b)/2 is equal to or greater than (ab)^1/2.

Answer 3

\begin{align} a > 0, b > 0, a + b < 2\sqrt{ab} &\implies a - 2\sqrt{ab} + b < 0 \\ &\implies (\sqrt a - \sqrt b)^2 < 0 \\ \end{align}

Which is a contradiction.

by the way

$a + b < 2 \sqrt{ab}\;$ implies $a^2-2ab+b^2 > 0\;$, not what you said.