Lagrange Multiplier, maximum value to show a equation

Question

I was doing a exercise and I don't understand a part of it.

1. Calculate the maximum value of $f (x, y, z) = \ln x + \ln y + 3 \ln z$ on the portion of the sphere $x^2+y^2+z^2=5r^2$ in the first octant. Use this result to show that $abc^3\leq 27\big((a+b+c)/5\big)^5$, for all $a,b,c >0$.

I got it do the first part of the problem, but i do not know what i should have to do in the second part, that ask to show that expression. ty I took a picture :)

Answer 1

Denote: $x=\sqrt{a}, y=\sqrt{b}, z=\sqrt{c}$. Then: $$\begin{align}f (x, y, z) = \ln x + \ln y + 3 \ln z&\le 5\ln r+3\ln \sqrt{3} \Rightarrow \\ \ln (xyz^3)&\le \ln (r^5\cdot \sqrt{27}) \Rightarrow \\ xyz^3&\le \sqrt{27}\cdot r^5\Rightarrow \\ \sqrt{abc^3}&\le \sqrt{27}\cdot \left(\sqrt{\frac{a+b+c}{5}}\right)^5 \Rightarrow \\ abc^3&\le 27\cdot \left(\frac{a+b+c}{5}\right)^5.\end{align}$$

Answer 2

Alt. hint (without Lagrange multipliers): $\;f(x,y,z)=\ln(xyz^3)\,$, and since $\,\ln\,$ is strictly increasing, the maximum value of $\,f\,$ is attained at the point which maximizes $\,xyz^3\,$. But by AM-GM:

$$ \frac{1}{27}x^2 y^2 z^6 = x^2 y^2 \frac{z^2}{3} \frac{z^2}{3} \frac{z^2}{3} \le \left(\frac{x^2 + y^2 + \frac{z^2}{3}+\frac{z^2}{3}+\frac{z^2}{3}}{5}\right)^5 = \left(\frac{x^2+y^2+z^2}{5}\right)^5 = r^{10} \tag{1} $$

Therefore $\,xyz^3 \le 3 \sqrt{3}\,r^5\,$ with equality iff $\,x=y=\dfrac{z}{\sqrt{3}}\,$.

Also, $\,(1)\,$ can be written as $\;abc^3 \,\le\, 27 \,\left(\dfrac{a+b+c}{5}\right)^5\,$ where $\,a=x^2, b=y^2, c=z^2\,$.