= Cube mean-gradient inequality
For the cube $Q=(0,L)^n$ and $u\in H^1(Q)$,
$$
\|u\|_{L^2(Q)}^2
\leq |Q|u_Q^2+\frac n2L^2\|Du\|_{L^2(Q)}^2,
\qquad
u_Q=\frac1{|Q|}\int_Qu.
$$
This follows from the pairwise-difference identity
$$
\int_Q|u-u_Q|^2=\frac1{2|Q|}\int_Q\int_Q|u(x)-u(y)|^2\,dx\,dy
$$
and a coordinate-by-coordinate path from $x$ to $y$. The <Cauchy-Schwarz inequality> bounds the squared sum of its $n$ increments by $n$ times the sum of their squares, while the one-dimensional <fundamental theorem of calculus> bounds each integrated increment by $L^{n+2}\|D_i u\|_2^2$. Division by $2|Q|=2L^n$ gives the stated constant.
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