See here http://www.artofproblemsolving.com/community/c6t243f6h1312835_nice_inequality_prove_4xyzgt2

We are given that $a,b,c$ are positive real numbers and $$ab+bc+ca+2abc=1$$ $$\Leftrightarrow \enspace ab+ca+2abc=1-bc \enspace \Leftrightarrow \enspace a = \frac{1-bc}{b+c+2bc}$$ Since, $a$ is positive we must have, $$a = \frac{1-bc}{b+c+2bc} \geq 0 \enspace \Leftrightarrow \enspace 1 \geq bc$$ So, the given inequality is equivalent to the following inequality $$\frac{4(1-bc)}{b+c+2bc} +b+c \geq 2$$ with the conditions, $$1 \geq bc \quad \quad b \geq 0 \quad \quad c \geq 0$$ Clearing the denominators the desired inequality is equivalent to, $$\frac{4(1-bc)}{b+c+2bc} +b+c \geq 2 $$ $$ \Leftrightarrow \enspace 4(1-bc) + (b+c)^2+ 2bc(b+c) \geq 2b+2c+4bc$$ $$ \Leftrightarrow \enspace b^2+c^2+4-6bc-2b-2c+2b^2c+2bc^2 \geq 0$$ $$ \Leftrightarrow \enspace b^2-2b+c^2-2c+2+ 2bc(b+c) -6bc +2 \geq 0$$ Now, using A.M.-G.M. inequality we have $2bc(b+c) \geq 4bc\sqrt{bc} = 4\sqrt{(bc)^3}$ Therefore, $$b^2-2b+c^2-2c+2 \enspace + \enspace 2bc(b+c) -6bc +2 \geq (b-1)^2+(c-1)^2 + 4(\sqrt{bc})^3 -6(\sqrt{bc})^2 +2$$ So, we just need to show that $$(b-1)^2+(c-1)^2 \enspace + \enspace 4(\sqrt{bc})^3 -6(\sqrt{bc})^2 +2 \geq 0$$ but this is equivalent to, $$(b-1)^2+(c-1)^2+ \enspace 4(\sqrt{bc})^3 -6(\sqrt{bc})^2 +2 = (b-1)^2+(c-1)^2 + \enspace 2(2\sqrt{bc}+1)(\sqrt{bc}-1)^2 \geq $$ which is obviously true. Hence, the desired inequality holds true. $\quad \quad \quad \quad \quad \quad \blacksquare$

We can easily solve this using substitution $a=\frac{z}{x+y} a=\frac{y}{x+z} c=\frac{z}{x+y} $ and just use some basic $AM-GM$ or Titu's