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

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 $a,b \leq 0$?

• Sorry, fixed it. I'm not entirely sure how $a,b \leq 0$ Follows from $(a-b)^2 < 0$. Nov 5, 2014 at 5:46
• Why you need to show $a,b\le 0$? Isn't $(a-b)^2<0$ a contradiction already? Nov 5, 2014 at 5:47
• In the calculation, you used without mentioning it the fact that $a+b\ge 0$. For without that you cannot go from $a+b\lt 2\sqrt{ab}$ to $a^2-2ab+b^2\lt 0$. Nov 5, 2014 at 5:50
• Possible duplicate of Proofs of AM-GM inequality
– sc_
Sep 29, 2018 at 17:17

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.

• It looks like the OP thinks that a proof by contradiction has to negate one of the assumptions or givens. In fact, proof 'by contradiction' is translated from latin (reductio ad absurdam) as reducing to an absurdity. The absurdity can be the negation of one of the assumptions, or it can be a statement that is known to be false, i.e. it is known to be false because it can be proven to be false separately or it is the negation of an axiom. To be pedantic you could state the negation of the known false statement, so you have an explicit inconsistency , p and not p.
– john
Jun 3, 2022 at 1:40
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.

\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}

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