Prove the following inequality holds: $\sqrt{(\log_3a^b +\log_3a^c)}+\sqrt{(\log_3b^c +\log_3b^a)}+\sqrt{(\log_3c^a +\log_3c^b)}\le 3\sqrt{6}$.

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

Prove the following inequality holds:

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

My solution:

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

[MATH]\le \sqrt{1+1+1}\sqrt{\log_3a^b +\log_3a^c+\log_3b^c +\log_3b^a+\log_3c^a +\log_3c^b}[/MATH]

[MATH]=\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}[/MATH]

[MATH]=\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})[/MATH]

[MATH]=\sqrt{3}\sqrt{(a+b+c)\log_3abc -(\log_3a^a+\log_3b^b+\log_3c^c})[/MATH]

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

[MATH]=\sqrt{3}\sqrt{27 -(a\log_3a+b\log_3b+c\log_3c})[/MATH]

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

[MATH]\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[/MATH]

[MATH]\therefore a\log_3a+b\log_3b+c\log_3c=3(3)=9[/MATH]

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

[MATH]\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*}[/MATH]

with equality when [MATH]a=b=c=3[/MATH].