Proving : $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge4(a+b+c)$ For $a,b,c > 0$ and $ab+bc+ca+2abc=1$, how to prove that:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge4(a+b+c) \, ?$$
 A: The condition is equivalent to $$\frac{a}{a + 1} + \frac{b}{b + 1} + \frac{c}{c + 1} = 1$$ then take $x = \frac{a}{a + 1}$, $y = \frac{b}{b + 1}$, $z = \frac{c}{c + 1}$ and the condition becomes $$x + y + z = 1$$ So we have $a = \frac{x}{1 - x} = \frac{x}{y + z}$ and similarly (by cyclic permutations) for other variables, on the other hand it is easily verified that $a = \frac{x}{y + z}$, $b = \frac{y}{z + x}$, $c = \frac{z}{x + y}$ satisfy the given condition, thus the condition is equivalent to saying that there exist positive real numbers $x,y,z$ such that $a = \frac{x}{y + z}$, $b = \frac{y}{z + x}$, $c = \frac{z}{x + y}$. Substituting this into the inequality we get that it is equivalent to $$\frac{y + z}{x} + \frac{z + x}{y} + \frac{x + y}{z} \geq 4 \cdot \left(\frac{x}{y + z} + \frac{y}{z + x} + \frac{z}{x + y}\right)$$ But this is easily proved because by Cauchy-Schwarz inequality we have $$(y + z)\left(\frac{x}{y} + \frac{x}{z}\right) \geq \left(\sqrt{x} + \sqrt{x}\right)^2 = 4x \Leftrightarrow \frac{x}{y} + \frac{x}{z} \geq \frac{4x}{y + z}$$ Analogously we have $$\frac{y}{x} + \frac{y}{z} \geq \frac{4y}{z + x} \text{ and } \frac{z}{x} + \frac{z}{y} \geq \frac{4z}{x + y}$$ Adding these last three inequalities yields the desired.
A: Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition does not depend on $u$, which says that it's enough to prove
our inequality for a maximal value of $u$, which happens for equality case of two variables.
Let $b=a$.
Hence, the condition gives $c=\frac{1-a}{2a}$, where $0<a<1$ and after these substitutions
we need to prove that 
$$\frac{1}{a}+\frac{2a}{1-a}\geq4\left(2a+\frac{1-a}{2a}\right)$$ or
$$(2a-1)^2\geq0.$$
Done!
