Let $x,y,z>0 $ and $xyz(x + y + z) = 1.$ Prove that$$(x+y)(y+z)\geq 2 $$ parmenides51 wrote: Find the least value of the expression $(x+y)(y+z)$, under the conditionthat $x,y,z$ are positive numbers satisfying the equation $xyz(x + y + z) = 1$.

Let $x,y,z>0 $ and $xyz(x + y + z) = 1.$ Prove that$$(2x+y)(y+2z)\geq 3\sqrt 3 $$Let $x,y,z>0 $ and $xyz(x + 2y + z) = 1.$ Prove that$$(x+y)(y+z)\geq \frac{3}{2}\sqrt{\frac{3}{2}}$$

41) $$\left ( a+b \right )\left ( a+c \right ) \ge 2\sqrt{abc\left ( a+b+c \right )}$$for $a,b,c \ge 0$