I am wondering if there is a more elementary proof of an inequality I found while reading a paper about minimal surfaces.
Let $f:\mathbb R^n\to\mathbb R$ be smooth, $Df(x)$ be its gradient, and $Q=\{x\in\mathbb R^n:|x^i|\le 1/2\}$ be the unit cube. Suppose we have a smallness bound on the "excess" functional (vertical part of the area): $$ \int_Q\sqrt{1+|Df|^2}dx-\int_Q dx\le E.\qquad (*) $$ Here, $E< 1$. Then this implies a small $W^{1,1}_0$ bound on the graph: $$ \int_Q|Df|dx\le 2\sqrt{E}.\qquad (**) $$ Is there a (more) elementary proof of $(**)$ (possibly modulo some constant independent of $E$)? Here is the proof I have seen:
Proof using $L^2$ duality:
Applying Cauchy-Schwarz inequality to each point $x$, we see that $$ \int_Q\sqrt{1+|Df(x)|^2}dx=\sup_{a(x)^2+|b(x)|^2\le 1}\int_Q\left(a(x)+b(x)\cdot Df(x)\right)dx, $$ where $a$ and $b$ range over smooth functions such that $a^2+|b|^2\le 1$. The sup is attained when we choose $a=1/\sqrt(1+|Df|^2)$ and $b=Df/\sqrt(1+|Df|^2)$. If we then make use of this in $(*)$, we get $$ \int_Q (a-1+b\cdot Df)dx\le E. $$ Choose $a=1-\tau$, where $\tau\in(0,1)$ is constant: $$ \int_Q (b\cdot Df)dx\le E+\tau. $$ Here, $b\in C^\infty(\mathbb R;\mathbb R^n)$ is arbitrary subject to $|b|^2\le 2\tau-\tau^2$ for $\tau\in(0,1)$, so taking the supremum in $b$ gives: $$ \sqrt{2\tau-\tau^2}\int_Q|Df|dx\le E+\tau. $$ Choosing $\tau=E$: $$ \int_Q|Df|dx\le\sqrt{\frac{2E}{1-E/2}}. $$ Applying $E<1$ to the denominator gives the result.
The function $g(x) = \sqrt{1+x^2}$ is convex, hence we can apply Jensen's inequality (note that the cube heas measure $1$) $$g\left( \int_Q|Df|\mathrm dx \right) \le \int_Q g(|Df|)\mathrm dx = \int_Q \sqrt{1+|Df|^2}\mathrm dx \le E + 1. $$ Rearranging yields $$ \int_Q|Df| \mathrm dx \le \sqrt{E^2 + 2\,E} $$ and for $E \in (0,1)$ this implies your inequality. Moreover, this inequality is sharp for $|Df| = \mathrm{const}$.