A.M - G.M equality. :D

From AM-GM inequality, we have $\frac{a}{b}+\frac{b}{a} \geq2$ and analogously $\frac{b}{c}+\frac{c}{b} \geq2$ , $\frac{c}{a}+\frac{a}{c} \geq2$ . Summing up these inequalities nets us: $\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b} +\frac{c}{a}+\frac{a}{c} \geq6$ , which has equality only at $\frac{a}{b}=\frac{b}{a}, \frac{b}{c}=\frac{c}{b}, \frac{c}{a}=\frac{a}{c}$ or equivalently $a^2=b^2=c^2$ , further simplifying to $a=b=c$ since we have $a,b,c \in \mathbb{R}^+$ . SInce our given conditions have this equality, this means we indeed have $a=b=c$ for these set of equations.

Substituting $a=b=c$ into the second equation nets us $3*\frac{a}{a^2}=\frac{3}{a}=4$ and consequently $\boxed{a=b=c=\frac{3}{4}}$

In Eqn 1, A.M - G.M Inequality.

a/b = b/a, b/c = c/b, c/a = a/c

=> a = b = c. Putting it in second eqn, we get 3/a = 4 => a = 3/4 = b = c