Inequality of logarithm function

[anemone]

Let the reals $a, b, c∈(1,\,∞)$ with $a + b + c = 9$.

Prove the following inequality holds:

$\sqrt{(\log_3a^b +\log_3a^c)}+\sqrt{(\log_3b^c +\log_3b^a)}+\sqrt{(\log_3c^a +\log_3c^b)}\le 3\sqrt{6}$.

 

[Albert1]

Let the reals $a, b, c∈(1,\,∞)$ with $a + b + c = 9$.

Prove the following inequality holds:

$\sqrt{(\log_3a^b +\log_3a^c)}+\sqrt{(\log_3b^c +\log_3b^a)}+\sqrt{(\log_3c^a +\log_3c^b)}\le 3\sqrt{6}$.

Spoiler

for:
$\log_3a^b >0,\log_3a^c>0,$
$\log_3b^a >0,\log_3b^c>0,$
$\log_3c^a >0,\log_3c^b>0,$
here $a, b, c∈(1,\,∞)$
if $a=b=c=3$ then equality holds
if $a=b=1,c=7$ then
$\sqrt{(\log_3a^b +\log_3a^c)}+\sqrt{(\log_3b^c +\log_3b^a)}+\sqrt{(\log_3c^a +\log_3c^b)}=\sqrt{(\log_37+\log_37)}<3\sqrt 6$
and $3\sqrt 6$ is a maximum value

 

[anemone]

Thanks Albert for participating!

My solution:

Spoiler

By applying the Cauchy-Schwarz inequality to the LHS of the intended inequality gives
\(\displaystyle \sqrt{(\log_3a^b +\log_3a^c)}+\sqrt{(\log_3b^c +\log_3b^a)}+\sqrt{(\log_3c^a +\log_3c^b)}\)

\(\displaystyle \le \sqrt{1+1+1}\sqrt{\log_3a^b +\log_3a^c+\log_3b^c +\log_3b^a+\log_3c^a +\log_3c^b}\)

\(\displaystyle =\sqrt{3}\sqrt{\log_3a^a +\log_3a^b+\log_3a^c+\log_3b^a +\log_3b^b+\log_3b^c+\log_3c^a +\log_3c^b+\log_3c^c-\log_3a^a-\log_3b^b-\log_3c^c}\)

\(\displaystyle =\sqrt{3}\sqrt{\log_3a^{a+b+c} +\log_3b^{a+b+c} +\log_3c^{a+b+c} -(\log_3a^a+\log_3b^b+\log_3c^c})\)

\(\displaystyle =\sqrt{3}\sqrt{(a+b+c)\log_3abc -(\log_3a^a+\log_3b^b+\log_3c^c})\)

\(\displaystyle =\sqrt{3}\sqrt{9\log_33^3 -(\log_3a^a+\log_3b^b+\log_3c^c})\) since \(\displaystyle a+b+c=9\ge 3\sqrt[3]{abc}\implies abc \le 3^3\)

\(\displaystyle =\sqrt{3}\sqrt{27 -(a\log_3a+b\log_3b+c\log_3c})\)

Now, if we're to study the nature of the function for \(\displaystyle f(a)=a\log_3 a\), we know it's a concave up and an increasing function, we could use the Jensen's inequality to figure out the minimum of \(\displaystyle a\log_3a+b\log_3b+c\log_3c\):

\(\displaystyle \frac{a\log_3a+b\log_3b+c\log_3c}{3}\ge \frac{a+b+c}{3}\log_3\left(\frac{a+b+c}{3}\right)=\frac{9}{3}\log_3\left(\frac{9}{3}\right)=3\)

\(\displaystyle \therefore a\log_3a+b\log_3b+c\log_3c=3(3)=9\)

Now we get the maximum of the LHS of the intended inequality as:

\(\displaystyle \begin{align*}\sqrt{(\log_3a^b +\log_3a^c)}+\sqrt{(\log_3b^c +\log_3b^a)}+\sqrt{(\log_3c^a +\log_3c^b)}&\le \sqrt{3}\sqrt{27 -(a\log_3a+b\log_3b+c\log_3c})\\& \le \sqrt{3}\sqrt{27 -9}\\&=\sqrt{3}\sqrt{18}\\&=\sqrt{9}\sqrt{6}\\&=3\sqrt{6}\end{align*}\)

with equality when \(\displaystyle a=b=c=3\).