<s>[INST] Prove that : $\frac{a+b+c+d}{a+b+c+d+f+g}+\frac{c+d+e+f}{c+d+e+f+b+g}>\frac{e+f+a+b}{e+f+a+b+d+g}$ 
Prove inequality for positive numbers:
  $$\frac{a+b+c+d}{a+b+c+d+f+g}+\frac{c+d+e+f}{c+d+e+f+b+g}>\frac{e+f+a+b}{e+f+a+b+d+g}$$

My work so far:
Lemma: If $x>y>0, t>z>0$, then $$\frac{x}{x+z}>\frac y{y+t}.$$
Proof. Really, $\frac zx< \frac ty\Rightarrow \frac{x+z}x<\frac{t+y}y.$
 [/INST] First, let's write the left side of the inequality as:
$L(c)=\dfrac{c+A}{c+B}+\dfrac{c+D}{c+E}$,
the derivative of $L(c)$ is:
$L'(c)=\dfrac{B-A}{(c+B)^2}+\dfrac{E-D}{(c+E)^2}$,
which is always positive for $c\geq 0$, because $B>A$ and $E>D$. Hence, L(c) is cresent for $c\geq 0$ and so, $L(c)>L(0)$ for $c\geq 0$, i.e.
$\dfrac{a+b+c+d}{a+b+c+d+f+g}+\dfrac{c+d+e+f}{c+d+e+f+b+g}>\dfrac{a+b+d}{a+b+d+f+g}+\dfrac{d+e+f}{d+e+f+b+g}$ $\cdots (1)$.
Now, working with the right side of the inequality,
$\dfrac{e+f+a+b}{e+f+a+b+d+g}=\dfrac{a+b}{e+f+a+b+d+g}+\dfrac{e+f}{e+f+a+b+d+g}$ $\cdots (2)$,
and by the positive condition in the numbers, it is not difficult to prove that the first term in $(1)$ is greater than the first term in $(2)$, and that the same happens with the second term of $(1)$ and $(2)$. Therefore, it is demostrated.
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