Equivalent norm in Sobolev space Let $\rho\in H^{1}(0,\pi)$ be a function, and consider the functional
$$
I(\rho)=\bigg(\int_{0}^{\pi}{\sqrt{\rho^2(t)+\dot\rho^2(t)}\,dt}\bigg)^2.
$$ 
I'm asking if it is equivalent to the norm 
$$
\lVert \rho \rVert_{H^1}=\lVert \rho \rVert_{L^2}+\lVert \dot\rho \rVert_{L^2}  
$$
on $H^{1}(0,\pi)$. Obviously $I(\rho)\leq \lVert \rho \rVert_{H^1}^2$, i'm asking if the other inequalities holds.

The functional $\sqrt{I(\rho)}$ cannot be equivalent to the norm $\|\rho\|_{H^1}$ becuase the latter cannot be equivalent to the norm
$$
\|\rho\|_{W^{1,1}}=\|\rho\|_{L^1}+\|\dot{\rho}\|_{L^1}\,.
$$ 
More precisely, due to an obvious fact
$$
\frac{1}{\sqrt{2}}(|\rho|+|\dot{\rho}|)\leqslant\sqrt{|\rho|^2+|\dot{\rho}|^2}
\leqslant(|\rho|+|\dot{\rho}|)
$$
holds the following inequality
$$
\frac{1}{\sqrt{2}}\|\rho\|_{W^{1,1}}\leqslant\sqrt{I(\rho)}
\leqslant\|\rho\|_{W^{1,1}}\,, 
$$
implying the equivalence of $\sqrt{I(\rho)}$ to the norm $\|\rho\|_{W^{1,1}}$ which is apparently weaker than the norm $\|\rho\|_{W^{1,2}}=\|\rho\|_{H^1}$.
