MHB Inequality of logarithm function

AI Thread Summary
The discussion centers on proving the inequality involving logarithmic functions of real numbers a, b, and c, constrained by the condition a + b + c = 9. The inequality to be proven is that the sum of the square roots of the logarithmic expressions is less than or equal to 3√6. Participants engage in deriving the proof and exploring the properties of logarithms to establish the validity of the inequality. The conversation highlights the mathematical techniques and reasoning necessary to tackle such inequalities. The proof ultimately confirms the stated inequality holds under the given conditions.
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Let the reals $a, b, c∈(1,\,∞)$ with $a + b + c = 9$.

Prove the following inequality holds:

$\sqrt{(\log_3a^b +\log_3a^c)}+\sqrt{(\log_3b^c +\log_3b^a)}+\sqrt{(\log_3c^a +\log_3c^b)}\le 3\sqrt{6}$.
 
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anemone said:
Let the reals $a, b, c∈(1,\,∞)$ with $a + b + c = 9$.

Prove the following inequality holds:

$\sqrt{(\log_3a^b +\log_3a^c)}+\sqrt{(\log_3b^c +\log_3b^a)}+\sqrt{(\log_3c^a +\log_3c^b)}\le 3\sqrt{6}$.
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
 
Thanks Albert for participating!

My solution:

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

$$\le \sqrt{1+1+1}\sqrt{\log_3a^b +\log_3a^c+\log_3b^c +\log_3b^a+\log_3c^a +\log_3c^b}$$

$$=\sqrt{3}\sqrt{\log_3a^a +\log_3a^b+\log_3a^c+\log_3b^a +\log_3b^b+\log_3b^c+\log_3c^a +\log_3c^b+\log_3c^c-\log_3a^a-\log_3b^b-\log_3c^c}$$

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

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

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

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

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

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

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

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

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

with equality when $$a=b=c=3$$.
 
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