6 Given that $\lambda$ is a positive real number. Find the maximum value of $\lambda$ such that for all positive real numbers $u, v, w$ satisfying the condition
$$u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v} \geqslant 1$$

we have
$$u+v+w \geqslant \lambda .$$
6. First, it is easy to observe that when $u=v=w=\frac{\sqrt{3}}{3}$, $u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v}=1$ and $u+v+w=\sqrt{3}$. Therefore, the maximum value of $\lambda$ does not exceed $\sqrt{3}$. Below, we prove that for all $u, v, w > 0$, and satisfying $u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v} \geqslant 1$, we have $u+v+w \geqslant \sqrt{3}$. By the AM-GM inequality and the Cauchy-Schwarz inequality, we have $\frac{(u+v+w)^{4}}{9}=\left(\frac{u+v+w}{3}\right)^{3} \cdot 3(u+v+w) \geqslant 3 u v w(u+v+w)=$ (uvw + vwu + wuv)(u+v+w) $\geqslant (u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v})^{2} \geqslant 1$. Therefore, $u+v+w \geqslant \sqrt{3}$. The equality holds if and only if $u=v=w=\frac{\sqrt{3}}{3}$. In summary, the maximum value of $\lambda$ is $\sqrt{3}$.