Logarithmic inequality for a>1

Question

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?

Answer 1

It surely isn't a nice solution, but you could proceed as follows: $$ log_{√a}(a+1)+log_{a+1}\sqrt a=\frac{\ln(a+1)}{\ln(\sqrt a)}+\frac{\ln(\sqrt a)}{\ln(a+1)}=\frac{2\ln(a+1)}{\ln(a)}+\frac{\ln(a)}{2\ln(a+1)} $$ What we now need to prove is: $$ \frac{2\ln(a+1)}{\ln(a)}+\frac{\ln(a)}{2\ln(a+1)}≥\sqrt6\iff \left( \frac{2\ln(a+1)}{\ln(a)}\right)^2+1≥\sqrt6\cdot\left( \frac{2\ln(a+1)}{\ln(a)}\right) $$ Since $a>1$ implies $\frac{2\ln(a+1)}{\ln(a)}>0$ by making the substitution $x=\frac{2\ln(a+1)}{\ln(a)}$ we need to prove: $$ x^2+1≥\sqrt6x\iff x^2-\sqrt6x+1≥0\iff\left(x-\frac{\sqrt6-\sqrt2}{2}\right)\left(x-\frac{\sqrt6+\sqrt2}{2}\right)≥0 $$ By the standard solution of quadratic equations. So we're done, if we can prove: $$ x-\frac{\sqrt6+\sqrt2}{2}≥0\iff x=\frac{2\ln(a+1)}{\ln(a)}≥\frac{\sqrt6+\sqrt2}{2}\iff2\ln(a+1)≥\frac{\sqrt6+\sqrt2}{2}\cdot\ln(a) $$ But from the standard inequality of the arithmetic and the quadratic mean, we have: $$ \frac{\sqrt6+\sqrt2}{2}<\sqrt{\frac{6+2}{2}}=2 $$ And since $a>1$ implies $\ln(a)>0$ we have: $$ \frac{\sqrt6+\sqrt2}{2}\cdot\ln(a)<2\ln(a)<2\ln(a+1) $$ Since $\ln(x)$ is strictly monotonically increasing. So we're done.

Answer 2

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 $a>1$

Answer 3

We can write this, with $b = \log_{\sqrt a}(a+1) > 0$ as $$b+\frac1b \ge \sqrt6 \iff b \ge \frac1{\sqrt2}+\sqrt\frac32 = c\iff 2\log(a+1) \ge c \log a \iff (a+1)^2 \ge a^c$$

which is obviously true as $a+1 > a$, and $2 > c$