From PDE Evans, 2nd edition: Chapter 8, Exercise 17:
Let $u,\hat{u} \in H_0^1(U)$ both be positive minimizers of the Dirichlet energy $$I[w] := \int_U |Dw|^2 \, dx,$$ subject to the constraint that $$\int_U w^2 \, dx = 1.$$ Suppose also that $u,\hat{u} > 0$ within $U$. Follow the hints to give a new proof that $$u \equiv \hat{u} \quad \text{in }U.$$ (Hint: Define $w:= \left(\frac{u^2+\hat{u}^2}2\right)^{1/2}$, $s := \frac{u^2}{u^2+\hat{u}^2}$ and $\eta := \frac{u^2+\hat{u}^2}2$; and show that $$|Dw|^2 = \eta\left|s \frac{Du}u+(1-s)\frac{D\hat{u}}{\hat{u}} \right|^2.$$ Deduce $$|Dw|^2 \le \eta \left(s\left|\frac{Du}u \right|^2+(1-s)\left|\frac{D\hat{u}}{\hat{u}} \right|^2 \right)=\frac 12|Du|^2+\frac 12|D\hat{u}|^2$$ and therefore $\frac{Du}u=\frac{D\hat{u}}{\hat{u}}$ almost everywhere.)
(Belloni-Kawohl, Manuscripta Math. 109 (2002), 229--231)
Okay, I actually proved almost all of the hint without any help (woo-hoo!). So I will not need to show any work here.
I do, however, have a question with the very last line of the hint. How can we prove $\frac{Du}u=\frac{D\hat{u}}{\hat{u}}$ a.e. given the relation $|Dw|^2 \le \frac 12|Du|^2+\frac 12|D\hat{u}|^2$? I did try to expand $|Dw|^2$, simplified a bit, then I got this: $$\left|\frac u{u^2+\hat{u}^2} Du + \frac{\hat{u}}{u^2+\hat{u}^2} D\hat{u} \right|^2 \le |Du|^2+|D\hat{u}|^2$$ but this doesn't seem to lead me anywhere.
Hint: If you integrate the last ineqality, you got
$$I[w] \le \frac 12 I[u]+ \frac 12 I[\hat u] = I[u]$$
But $u$ is a minimum of $I$, so all the inequalities are equalities.