Logarithmic inequality for a>1 Is $\log_{\sqrt a}(a+1)+\log_{a+1}\sqrt a\ge \sqrt6$ always true for $a>1$?
What is the approach? Do we check the first a's and then form a induction hypothesis?

rewrite your inequality in the form
$\frac{\ln(a+1)}{\ln(\sqrt{a})}+\frac{\ln(\sqrt{a})}{\ln(a+1)}=$
$\frac{2\ln(a+1)}{\ln(a)}+\frac{\ln(a)}{2\ln(a+1)}\geq 2$
by the theorem that stated
$\frac{x}{y}+\frac{y}{x}\geq 2$ iff $(x-y)^2\geq 0$
for $x,y>0$
if we define $f(a)=\frac{2\ln(a+1)}{\ln(a)}+\frac{\ln(a)}{2\ln(a+1)}$ we get for the first derivative 
$f'(a)=-1/2\,{\frac { \left( \ln  \left( a \right) -2\,\ln  \left( a+1
 \right)  \right)  \left( \ln  \left( a \right) +2\,\ln  \left( a+1
 \right)  \right)  \left( \ln  \left( a \right) a-a\ln  \left( a+1
 \right) -\ln  \left( a+1 \right)  \right) }{ \left( a+1 \right) 
 \left( \ln  \left( a \right)  \right) ^{2}a \left( \ln  \left( a+1
 \right)  \right) ^{2}}}
$
and $f'a)<0$ for all $a>1$ thus we get by
$\lim_{a \to \infty}f(a)=\frac{5}{2}$ and we get $f(a)\geq \frac{5}{2}$ for $a>1$
