Let $u>0$ be a solution of the following equation $$\begin{cases}-[|u'(x)|^{p-2}u'(x)]'=\lambda_1\cdot u(x)^{p-1}&,x \in (a,b)\\u(a)=u(b)=0 \end{cases},$$ where $\lambda_1$ is the minimum of the Rayleigh quotient and $p>1.$
Prove that the function $\log u(x)$ is concave on $(a,b).$
It is easy to see that $\lambda_1>0$. From the first equation, one has $$ -(p-2)|u'(x)|^{p-3}\frac{u'(x)}{|u'(x)|}u''(x)u'(x)-|u'(x)|^{p-2}u''(x)=\lambda_1u(x)^{p-1} $$ or $$ -(p-1)|u'(x)|^{p-2}u''(x)=\lambda_1u(x)^{p-1}. $$ from which it is easy to see $$ u''(x)=-\frac{\lambda_1}{p-1}\frac{u(x)^{p-1}}{|u'(x)|^{p-2}}<0. \tag{1} $$ So by (1), one has \begin{eqnarray*} (\ln u(x))''&=&\left(\frac{1}{u(x)}u'(x)\right)'\\ &=&-\frac{1}{u(x)^2}|u'(x)|^2+\frac{1}{u(x)}u''(x)\\ &=&-\frac{1}{u(x)^2}|u'(x)|^2+\frac{1}{u(x)}u''(x)\\ &<&0. \end{eqnarray*} Namely $\ln u(x)$ is concave.