Curious inequality: $(1+a)(1+a+b)\geq\sqrt{27ab}$ I was recently trying to play with mean inequalities and Jensen inequality. My question is, if the following relation holds for any positive real numbers $a$ and $b$
$$(1+a)(1+a+b)\geq\sqrt{27ab}$$
and if it does, then how to prove it. By AM-GM inequality we could obtain $1+a\geq\sqrt{4a}$ and $1+a+b\geq\sqrt[3]{27ab}$, so by putting this together we obtain$$(1+a)(1+a+b)\geq\sqrt{4a}\sqrt[3]{27ab},$$but this is slightly different from what I wanted. It's probably true that for all positive $a$ and $b$ there is $\sqrt{4a}\sqrt[3]{27ab}\geq\sqrt{27ab}$, but could there be equality?  
Thanks a lot. 

Use AM-GM:
$$\frac{1}{2}+\frac{1}{2}+a\ge3\sqrt[3]{\frac{a}{4}}\\
\frac{1}{2}+\frac{1}{2}+a+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}\ge6\sqrt[6]{\frac{ab^3}{108}}\\\therefore(1+a)(1+a+b)\ge\left(3\sqrt[3]{\frac{a}{4}}\right)\left(6\sqrt[6]{\frac{ab^3}{108}}\right)=\sqrt{27ab}$$
Equality holds iff $\frac{1}{2}=a=\frac{b}{3}$.
