Inequality on product of two positive numbers This question is linked to this other question about a proof of the AM-GM inequality.
All that I still don't understand, because I don't know how to prove, is the Lemma, by which if $a, b, c \in \mathbb{R^+}$ and $a > b$, and if we can choose a $c$ such that $a-c \geq b+c$, then $(a-c)(b+c) > ab$.
This is not so intuitive, but I guess it can be proved with some elementary calculus. Anyway, I was asked not to use calculus to prove it, but only algebra, and the only thing I managed to do was:
$$(a-c)(b+c) > ab \rightarrow ac-bc > c^2 \rightarrow a-b > c$$
But didn't we have to obtain something like $a-b > 2c$, by hypothesis?
Is there any other trick which I can exploit this inequality with?
 A: The Lemma (slightly modified) says
that if $a$, $b$, $c$ are nonnegative real numbers such that $a\geq b$ and $a-c\geq b+c$,
then $(a-c)(b+c)\geq ab$.
You are reasoning in a wrong direction. In the opposite direction you get
$a-c\geq b+c$ $\,\Longleftrightarrow\,$ $a-b\geq 2c$
$\,\Longrightarrow\,$ $a-b\geq c$ $\,\Longrightarrow\,$ $(a-c)(b+c)\geq ab\,$.
I wonder why anybody would go proving the inequality $\sqrt{ab}\leq\frac{1}{2}\!(a+b)$
using the Lemma you quote;
it is not as if the Lemma offers any insight into the nature of the inequality,
on the contrary, it obfuscates it.
The way to really understand the inequality is to look at the identity
$$
\Bigl(\frac{a+b}{2}\Bigr)^2 \!-\, ab ~=~ \Bigl(\frac{a-b}{2}\Bigr)^2~,
$$
which also provides a lower bound
for the gap between $\sqrt{ab}$ and $\frac{1}{2}\!(a+b)$ (assuming $a+b>0$),
$$
\tfrac{1}{2}\!(a+b)-\sqrt{ab}
    ~=~ \frac{\tfrac{1}{4}\!(a-b)^2}{\tfrac{1}{2}\!(a+b)+\sqrt{ab}}
    ~\geq~ \frac{(a-b)^2}{4\mspace{1mu}(a+b)}~,
$$
where the equality holds iff $a=b$.
Added a little later. $~$Looking at the
other question about a proof of the
AM-GM inequality,
I understand the need for the Lemma:
there the arithmetic and geometric means are of any number of positive numbers,
not of only two.
The derivation above of $(a-c)(b+c)\geq ab$
from $a\geq b$ and $c\geq 0$ and $a-c\geq b+c$
is done by shuffling the symbols, obeying certain rules.
We want to actually see what this result means,
so assume that $a>b$ (the case $a=b$ is quite uninteresting)
and consider the diagram of the function $(a-x)(b+x)$
on the interval $-b\leq x\leq a\,$:
$\hspace{80pt}$
Just a glance at the diagram convinces us that $(a-x)(b+x)>ab$ for $0<x<a-b$,
and you have actually proved it from $(a-x)(b+x)-ab=(a-b-x)\mspace{1mu}x\mspace{2mu}$
(only you wrote $c$ instead of $x$).
This also can be visualized: just lift the origin $(0,0)$ to the point $(0,ab)$,
and then the diagram above (well, its part for $0\leq x\leq a-b$)
will look like this:
$\hspace{80pt}$
Therefore, if the inequality $(a-x)(b+x)>ab$ is what we want,
there is no need to restrict $x$ to the interval $0<x\leq\frac{1}{2}\!(a-b)$.
$\newcommand{\amean}{\overline}
$Let us look at how the Lemma is used in the proof of the inequality
$$
\bigl(a_1\cdots a_n\bigr)^{1/n} \:\leq\: \frac{1}{n}\!\bigl(a_1+\cdots+a_n\bigr)~,\tag{1}
$$
where $n\geq 1$ and $a_1,\ldots,a_n>0$,
and the equality holds iff $a_1=\cdots=a_n$.
If we already have $a_1=\cdots=a_n$ to start with, then $(1)$ is an equality.
$\qquad$Suppose that not all $a_k$ are equal to each other,
and write $\amean{a}=\frac{1}{n}\!(a_1+\cdots+a_n)$.
There exist indices $i\neq j$ such that $a_i<\amean{a}<a_j$.
We increase $a_i$ to $a_i'<a_j$
and at the same time decrease $a_j$ to $a_j'>a_i$
so that $a_i'+a_j'=a_i+a_j$,
and then we replace $a_i$ with $a_i'$ and $a_j$ with $a_j'$
(leaving all $a_k$ with $k\neq i,j$ as they are);
then the right hand side of $(1)$ (which is $\amean{a}$) remains unchanged
while the left hand side strictly increases (here we used the Lemma).
Observe that no matter where in the interval $(a_i,a_j)$ we choose $a_i'$,
its companion $a_j'$ also lies in the interval.
Now the idea is to move
one of $a_i$, $a_j$ to $\amean{a}$.
I suspect that the official proof of the inequality GM $\leq$ AM which uses the Lemma
(I have not seen the proof) always chooses to move
that one of $a_i$, $a_j$ that is closest to $\amean{a}$
-- which would explain the restriction $a-c\geq b+c$ in the Lemma.
Actually we can always choose to increase $a_i$ to $\amean{a}$
or always choose to decrease $a_j$ to $\amean{a}$,
since it all comes to the same thing: after the adjustment
we get $a_i'+a_j'=a_i+a_j$ and $a_i'a_j'>a_ia_j$, 
and at least one of $a_i'$, $a_j'$ equals $\amean{a}$
while $a_i\neq\amean{a}$ and $a_j\neq\amean{a}$.
If after the first adjustment the new $a_k$'s
(we silently rename $a_i'$, $a_j'$ to $a_i$, $a_j$)
are still not all equal to each other and hence not all equal to $\amean{a}$,
we adjust again, then perhaps again$\ldots\,$
until after at most $n-1$ adjustments we will have all $a_k$'s equal to $\amean{a}$
and $(1)$ will become equality.
Since the right hand side of $(1)$ remained unchanged
through all the adjustments
while each adjustment strictly increased the left hand side of $(1)$,
we conclude that we must have started with
a strict inequality GM $<$ AM.
A: We have 
$$a-c\ge b+c\iff a-b\ge 2c.$$
Hence, we can say $a-b\gt c$ because $a-b\ge 2c\gt c$.
