Mathematics Proof - prove that $p+q > \sqrt{4pq}$ I want to ask about a question I found in a mathematics textbook today in my lecture.
This was the question:

Prove that for any positive numbers p and q:
$p+q > \sqrt{4pq}$

Now, this is what I did.
I worked backwards to find an expression $f(x) > 0$
$$p+q > \sqrt{4pq}$$
$$(p+q)^2 > 4pq$$
$$(p+q)^2 - 4pq > 0$$
So, I looked at the expression $(p+q)^2$.
I know that adding two producing numbers and squaring them means that the result will be greater than zero, so $(p+q)^2 > 0$
Now, I also noted that $(p+q)^2$ or better shown as $p^2 + 2pq + q^2$ is greater than $4pq$ so I thought in the end, this must mean that
as
$$(p+q)^2 > 4pq$$
I can show hence that
$$p+q > \sqrt{4pq}$$
However, the book gives this very strange answer. I will attach it as a picture for reference.


$(p+q)^2 = p^2 + 2pq + q^2 = (p-q)^2 + 4pq$
$(p-q)^2 ≥ 0 $ since it is a square, so $(p+q)^2 ≥ 4pq$
$p > 0, q > 0 \Rightarrow p + q > 0 \Rightarrow p+q > \sqrt{4pq}$

I'm confused. What is the book trying to say, and is my answer a valid proof?
many thanks.
 A: You wrote in your answer: I also noted that $(p+q)^2$ is greater than $4pq$. But that is actually not obvious. To prove that you need to say that $(p-q)^2>0$ (since it's a square), hence $p^2 -2pq + q^2>0$ and adding $4pq$ to both parts we get the desired inequality.
A: 
Now, I also noted that $(p+q)^2$ or better shown as $p^2+2pq+q^2$ is greater than $4pq$ so I thought in the end, this must mean that as $(p+q)^2>4pq$. 

This part of your proof makes no sense, and is where your proof is lacking.
Here's an explanation of the book's proof:
$$
\begin{align}
(p-q)^2 &\ge 0\tag{all squares are positive}\\
(p-q)^2 + 4pq &\ge 4pq\\
(p+q)^2 &\ge 4pq \tag{rearrange the left hand side}\\\
p+q &\ge \sqrt{4pq} \tag{take square roots}
\end{align}
$$
A: You did not prove that $(p+q)^2\geq4pq$.
If $(p+q)^2>0$ it not gives that $(p+q)^2\geq4pq$.
The book tries to say that from $(p+q)^2\geq4pq$ follows $p+q\geq\sqrt{4pq}$ if $p+q\geq0$.
For example $(-3)^2>4$ is true, but $-3>\sqrt{4}$ is wrong.
A: Book is right. You say:
$ p^2+q^2+2pq $ is greater than 4pq
From what you know this? It is the same you must prove.
and 
  $(p-q)^2 \geq 0 $ proves it.
A: I think you should use a simple inequality relation here,
Arithmetic Mean≥ Geometric Mean(AM≥GM)
Now consider two numbers $p$ and $q$
AM= ${p+q}/2$
GM=$\sqrt{pq}$
Since, AM≥GM
That's why,
${p+q}/2$≥$\sqrt{pq}$
OR,
${p+q}≥2\sqrt{pq}$
OR,
${p+q}≥\sqrt{4pq}$
A: You haven't made the connection that $(p+q)^2\ge4pq\implies (p-q)^2\ge0$ obvious. Your proof is very close, but obfuscates this connection, which is the heart of the proof. 
A: I did not actually answer the question. 
Considering this all again, we know that $(p-q)^2 > 0$ because it is a square number, hence 
$$p^2 + q^2 - 2pq > 0$$
so $$p^2 + q^2 + 2pq > 4pq$$
$$(p+q)^2 > 4pq$$
$$p+q > \sqrt{4pq}$$
A: Counter-example
The book got it right, but your statement is in fact false.
The book proved $p+q\ge\sqrt{4pq}$
You wrote $p+q\gt\sqrt{4pq}$
It turns out if $p=q$ the equation becomes $p+p\gt\sqrt{4pp}$ or $2p\gt2p$ which is false.  It turns out your statement is true only for $p\ne q$.
