Question about Lagrange multiplier and maximum point Find the maximum of $\log{x}+\log{y}+3\log{z}$ on portion of the sphere $x^2 + y^2 + z^2 =5r^2$ 
where $x,y,z>o $
I found that maximum is $5\log{r} + 3\log{\sqrt{3}}$ at $(r,r,3\sqrt{3})$
And Use this result to prove that for real positive $a, b, c$, 
$abc^3$ is less than or equal to $27(a+b+\frac{c}{5})^5$ 
Can anyone help me prove this inequality given that we have a maximum?

There are two typos:
The maximum happens at $(r,r,\sqrt{3}r)$, and the right hand side of the inequality should be $27(\frac{a+b+c}{5})^5$.
Suppose $a=x^2,b=y^2,c=z^2$. Use the fact that $\log{x}+\log{y}+3\log{z}\leq 5\log{r} + 3\log{\sqrt{3}}$ when $x,y,z$ satisfies the sphere equation. 
So
$$\log{xyz^3}\leq\log{3\sqrt{3}r^5}\\
\Rightarrow \log{x^2y^2z^6}\leq \log{27r^{10}}\\
\Rightarrow \log{abc^3}\leq \log{27(\frac{a+b+c}{5})^5}\\
\Rightarrow abc^3\leq 27(\frac{a+b+c}{5})^5$$
I used the fact that $r^2=\frac{x^2+y^2+z^2}{5}$
Some log properties I used:
$$\log{a+b}=\log{a}+\log{b}\\
\log{a^r}=r\log{a}$$
