I came across the following theorem, while studying "A First Course in Real Analysis" by Berberian Sterling.
In an ordered field, if $a, b, c \geq 0$ and $a \leq b+c$, then $$ {{a}\over{1+a}} \leq {{b}\over{1+b}} + {{c}\over{1+c}}$$
Here is my proof for this problem.
I first start by proving the following,
[ T1 ]: if $x \leq y$ and $x, y > 0$ then ${{x}\over{1+x}} \leq {{y}\over{1+y}}$ $$ x \leq y \implies {{1}\over{x}} \geq {{1}\over{y}} \implies 1+{1 \over x} \geq 1+{1 \over y} \implies {x \over 1+x} \leq {y \over 1+y} $$
Now considering the right hand side of the statement to be proved $$ {b \over 1+b} + {c \over 1+c} = {b+c+2bc \over 1+b+c+bc} $$
From T1, since $a \leq b+c \implies a \leq b+c+bc$ and ${bc \over 1+b+c+bc} > 0$, $$ {b+c+2bc \over 1+b+c+bc} = {b+c+bc \over 1+b+c+bc} + {bc \over 1+b+c+bc} \geq {b+c+bc \over 1+b+c+bc} \geq {a \over 1+a} $$
Thus, ${b \over 1+b} + {c \over 1+c} \geq {a \over 1+a}$.
Assuming I have not made any mistakes in the above proof, I would love to know if there is an alternative (and possibly more elegant) way to prove this?
First, observe that $f(x) = \frac{x}{1+x}$ is an increasing function.
Thus, if $a \leq b$, then $f(a) \leq f(b) \leq f(b) + f(c)$. Similarly if $a \leq c$.
Now suppose $b \leq a$ and $c \leq a$. Then
$$ \frac{a}{a+1} \leq \frac{b}{a+1} + \frac{c}{a+1} \leq \frac{b}{b+1} + \frac{c}{c+1} $$
As an aside, to prove my observation, let $P$ be the statement "$f(x)$ is a strictly increasing function on $x \geq 0$". I'm lazy and I already know how to use calculus to prove P, so I don't want to figure out how to do it with algebra (although it's probably pretty easy). Here is my favorite proof method:
As a function on the real numbers, $f(0) = 0$ and $f'(x) = 1/(1+x)^2 > 0$, therefore $P$.
$P$ is a semialgebraic (i.e. it can be written in the first-order language of ordered fields). Therefore if it's true in one real closed field (e.g. the real numbers), it's true in all real closed fields. (because the theory of real closed fields is complete)
Finally, every ordered field can be embedded in a real closed field. Therefore, $P$ is true in all ordered fields.