1. ## Prove that

$a\in R^{+},b\in R^{+}$
if : $a+b+c=1$
Prove that :
$
\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\leq 2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})
$

2. Shouldn't it be: $c\in \mathbb{R}^+$ too?

We have:
$\sum\frac{1+a}{1-a}=\sum\left(\frac{1+a}{1-a}-1\right)+3=\sum\left(\frac{2a}{1-a}\right)+3=\sum\left(\frac{2a}{b+c}\right)+3$ - all the sums are cyclic-

Thus: $\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\leq{2\cdot\left(\frac{a}{b}+\frac{b}{c}+\frac{c }{a}\right)}$ iff $\frac{3}{2}\leq \sum\left(\frac{a}{b}\right) - \sum\left(\frac{a}{b+c}\right)$ (1)

Now: $\text{RHS} = \sum\left(\frac{a}{b}-\frac{a}{b+c}\right) = \sum\left( \frac{a}{b}\cdot (1-\tfrac{b}{b+c}) \right)=\sum\left( \frac{a}{b}\cdot \frac{c}{b+c} \right)=\sum\left( \frac{ac}{b^2+bc}\right)$ (2)

Let $x=\frac{a}{b};y=\frac{b}{c};z=\frac{c}{a}$, then (2) is equivalent to: $RHS=\sum\left(\frac{x}{y+1}\right)$ -remember, it's cyclic-

By Cauchy-Schwarz: $\text{RHS}\cdot \left(\sum (x\cdot(y+1))\right)\geq \left(\sum x \right)^2$ (3)

Let $q = \sum x$ then : $\sum (x\cdot(y+1))= q + \sum (xy) \leq q + \tfrac{q^2}{3}$ $(*)$

Thus by (3) we have: $\text{RHS}\geq \frac{q}{1+\tfrac{q}{3}}=\tfrac{3q}{3+q}$

But since $q\geq 3$ by AM-GM inequality -remember $x\cdot y \cdot z = 1$ -, it follows that $\tfrac{3q}{3+q}\geq\frac{3}{2}$ which proves (1) $\square$

$(*)$ Since $(\sum x)^2 = \sum x^2 + 2\sum xy$ and $\sum x^2 \geq \sum xy$