Is this inequality true? If yes, for what functions? Let $B=B(0,1)\subset \mathbb R^2$. Let $u$ be a radially symmetric differentiable function on $B$ and $v=Ax+b$ be a linear function where $A$ is a $2\times 2$ matrix satisfies $A=-A^T$, and $b=(b_1,b_2)$ is any constant. 
Define for $a=(a_1,a_2)\in\mathbb R^2$, $|a|=\sqrt{a_1^2+a_2^2}$.
I want to prove
$$
\int_B |\nabla u|\,dx\leq \int_B|\nabla u-Ax-b|\,dx. \tag 1
$$
I think this inequality probability is not true but I can not find a counterexample... 
Please help me to find a counterexample and if possible, what kind of assumptions should I add to $u$ so that equation $(1)$ is true?
Thank you!
 A: I was sceptical at the beginning, but the formula turns out to be true at the end. Here is the $n$-dimensional variant.
Let $N$ denote the outer unit normal for the sphere $S_r=\{|x|=r\}$, $0<r\le 1$. We have
$$
\int_{S_r}|\nabla u|\,ds_r=\left|\int_{S_r}\nabla u\cdot N\,ds_r\right|=
\left|\int_{S_r}(\nabla u+Ax+b)\cdot N\,ds_r\right|\le\int_{S_r}|\nabla u+Ax+b|\,ds_r.
$$
Here we used that $Ax\bot N$ and the flux of the constant field $b$ over the sphere is zero (from the divergence theorem since $b^Tx$ is harmonic or can be seen directly). Finally, the integration wrt $r$ from $0$ to $1$ gives
$$
\int_B|\nabla u|\,dx\le\int_B|\nabla u+Ax+b|\,dx.
$$
Generalization: for any harmonic function $v$ and for any vector field $F(x)\bot x$ in the unit ball $B$ we have
$$
\int_B|\nabla u|\,dx\le\int_B|\nabla u+F(x)+\nabla v|\,dx.
$$
A: Since $u$ is radially symmetric, we can write $u(x) = f(\lvert x\rvert)$ with $f \colon [0,1) \to \mathbb{R}$, and since $u$ is assumed differentiable, $f$ must be differentiable too. Let's assume that $f$ is continuously differentiable for simplicity.
For every $x \neq 0$, we can write
$$b = \underbrace{\frac{\langle b, x\rangle}{\lvert x\rvert^2}\cdot x}_{p(x)} + w(x)$$
with $w(x) \perp x$. Then we can split
$$v(x) = \nabla u(x) - Ax - b = (\nabla u(x) - p(x)) - (Ax + w(x)).$$
Since $(Ax + w(x)) \perp (\nabla u(x) - p(x))$, we have
$$\lvert \nabla u(x) - p(x)\rvert \leqslant \lvert \nabla u(x) - Ax - b\rvert.\tag{1}$$
Now, for $0 \leqslant \varphi < \pi$, look at
$$\int_{-1}^1 \lvert\nabla u(t\cdot e_\varphi) - p (t\cdot e_\varphi)\rvert\,dt,$$
where $e_\varphi = (\cos \varphi, \sin \varphi)$. Let us fix $\varphi$ and write $g(t) = \nabla u(t\cdot e_\varphi),\; h(t) = p(t\cdot e_\varphi)$ for $t\in (-1,1)$ for a shorter notation. Since
$$\nabla u(x) = \frac{f'(\lvert x\rvert)}{\lvert x\rvert}\cdot x$$
we have $g(-t) = -g(t)$. And we have $h(-t) = h(t)$, so
$$\langle g(t),h(t)\rangle > 0 \iff \langle g(-t),h(-t)\rangle < 0.$$
Hence
\begin{align}
\int_{-1}^1 \lvert g(t) - h(t)\rvert\,dt
&= \int_0^1 \lvert g(t) - h(t)\rvert + \lvert g(-t) - h(-t)\rvert\,dt\\
&= \int_0^1 \bigl\lvert\lvert g(t)\rvert \pm \lvert h(t)\rvert\bigr\rvert + \bigl\lvert\lvert g(-t)\rvert \mp \lvert h(-t)\rvert\bigr\rvert\,dt\\
&\geqslant \int_0^1 \lvert g(t)\rvert \pm \lvert h(t)\rvert + \lvert g(-t)\rvert \mp \lvert h(-t)\rvert\,dt\\
&= \int_0^1 \lvert g(t)\rvert + \lvert g(-t)\rvert\,dt\\
&= \int_{-1}^1 \lvert g(t)\rvert\,dt,
\end{align}
so we see
$$\int_B \lvert \nabla u(x) - p(x)\rvert\,dx \geqslant \int_B \lvert \nabla u(x)\rvert\,dx,$$
which together with $(1)$ proves the desired inequality.
A: This is only a partial answer, but I'm out of time to work on it further just now, and so I'm putting it up in hopes that someone will be able to fill in the missing step in the comments.  It's ugly and coordinate-based, and it's only got a physicist's level of rigor;  but I suspect that the inequality is in fact true.  The one missing step is in boldface below.

As noted in the comments, we can choose coordinates such that $\vec{b} = (b,0)$.  We can also write 
$$
A = \begin{bmatrix} 0 & a \\ -a & 0 \end{bmatrix}.
$$
In polar coordinates $\{r, \theta\}$, the vectors in question are then
$$
\vec{\nabla} u = \frac{\partial u}{\partial r} \hat{r}, \qquad A \vec{x} = a r \hat{\theta}, \qquad \vec{b} = b (\cos \theta \hat{r} - \sin \theta \hat{\theta}).
$$
Brute-forcing the integrand, then, we have after some algebra
\begin{align}
I \equiv \left| \vec{\nabla} u - A \vec{x} - \vec{b} \right| &= \left[ \left(\frac{\partial u}{\partial r} \right)^2 + b^2 + a^2 r^2 - 2 b \left( \left(\frac{\partial u}{\partial r} \right) \cos \theta + a r \sin \theta  \right)\right]^{1/2} \\
&= \left[ C^2 + b^2  - 2 b C \cos (\theta - \theta_0)\right]^{1/2}
\end{align}
where 
$$
C^2 = \left(\frac{\partial u}{\partial r} \right)^2 + a^2 r^2
$$
and $\theta_0$ is determined in terms of the above quantities as well.  (It will turn out to be unimportant in just a second.)
This will need to be integrated over $r \in [0,1]$ and $\theta \in [0, 2\pi]$.  In particular, when we do the integral with respect to $\theta$, we get
$$
\int_0^{2\pi} I \, d\theta = \int_{\theta_0}^{\theta_0 + 2 \pi} \left[ C^2 + b^2  - 2 b C \cos (\theta - \theta_0) \right]^{1/2} \, d \theta = \int_{0}^{ 2 \pi} \left[ C^2 + b^2  - 2 b C \cos \theta' \right]^{1/2} d\theta'
$$
(using the periodicity of the integrand in the first step and defining $\theta' = \theta - \theta_0$ in the second.
This latter integral can be expressed in terms of complete elliptic integral of the second kind:
$$
\int_0^{2\pi} I \, d\theta = 4 |C - b| \, E \left( - \frac{4 b C}{(C- b)^2} \right).
$$
It would therefore suffice to prove that this quantity is greater than $2 \pi |C|$ for all values of $b$;  if this were true, then we would have
$$ 
\int_0^{2\pi} I \, d\theta \geq \int_0^{2\pi} |C| \, d \theta \geq \int_0^{2\pi} \left| \frac{\partial u}{\partial r} \right| \, d \theta,
$$
and integrating both sides with respect to $r$ would yield the desired inequality.  Equality would hold only when $a = b = 0$.
