Math Proof Help

Question

So I am supposed to add some condition to the original proposition to make it true but I do not know what condition I need to add.

Original Proposition: If $x$ and $y$ are real numbers and $xy>0$, then: $$(x+y)/2≥√xy$$

Proof: Let us assume the hypothesis to be true. Adding some condition to the original proposition we get: $x-y≥0$

By algebra we get: $$x-y≥0$$ $$(x-y)^2≥0$$ $$x^2-2xy+y^2≥0$$ $$x^2+2xy+y^2≥4xy$$ $$(x+y)^2≥4xy$$ $$x+y≥2√xy$$ $$(x+y)/2≥√xy$$

I thought that I needed to add the condition that $x≥y$ but this isn't the case because let's say the $x=-1$ and $y=-1$ it would be false. I then thought that $x$ and $y$ need to be positive integers but then I don't know how to use that to get $x-y≥0$

Answer 1

You don't actually need that $x-y \geq 0$, as $(x-y)^2 \geq 0$ no matter the value of $x-y$. What you do need is that $x+y \geq 0$, as when you take the square root in the second to last step you need to know that $x+y$ is the positive one (and not $-(x+y)$).

When $xy > 0$ this is equivalent to $x,y \geq 0$.

Answer 2

One: (both $x,y$ must be positive)

Notice $$( \sqrt{x} - \sqrt{y})^2 = x - 2 \sqrt{xy} + y \geq 0 \implies \frac{x+y}{2} \geq \sqrt{xy} $$

Another solution follows from young inequality: If $x,y$ are positive and $m +n = 1 $, then

$$ x^my^n \leq xm + yn $$

Now, if we put $m = n = \frac{1}{2} $, then

$$ x^{1/2}y^{1/2} = \sqrt{xy} \leq \frac{x}{2} + \frac{y}{2} = \frac{x + y}{2}$$