Prove: $\log_2(x)+\log_3(x)+\log_5(x)>9\log_{30}(x)$ 
Prove for all $x>1$
$\log_2(x)+\log_3(x)+\log_5(x)>9\log_{30}(x)$

So what I did was:
\begin{align}
&\frac{\ln(x)}{\ln(2)}+\frac{\ln(x)}{\ln(3)}+\frac{\ln(x)}{\ln(5)}>9\frac{\ln(x)}{\ln(30)}
\\[6px]
&\frac{\ln(x)}{\ln(2)}+\frac{\ln(x)}{\ln(3)}+\frac{\ln(x)}{\ln(5)}>9\frac{\ln(x)}{\ln(2\cdot3\cdot5)}=9\frac{\ln(x)}{\ln(2)+\ln(3)+\ln(5)}
\\[6px]
&\frac{\ln(x)}{\ln(2)}+\frac{\ln(x)}{\ln(3)}+\frac{\ln(x)}{\ln(5)}\ge3\frac{3\ln(x)}{\sqrt[3]{\ln(2)\ln(3)\ln(5)}}
\\[6px]
&\!\ln(2)+\ln(3)+\ln(5)\ge3\sqrt[3]{\ln(2)\ln(3)\ln(5)}
\\[6px]
&\frac{3\ln(x)}{\sqrt[3]{\ln(2)\ln(3)\ln(5)}}>9\frac{\ln(x)}{\ln(2)+\ln(3)+\ln(5)}
\end{align}
I think this should prove the inequality but I'm not really sure how to formulate it properly if it was in an exam, any tips how to conclude? (if my proof is correct)
 A: Consider the left hand side:
$$\log_2{x}+\log_3{x}+\log_5{x} = \frac{\log_{30}{x}}{\log_{30}{2}}+\frac{\log_{30}{x}}{\log_{30}{3}}+\frac{\log_{30}{x}}{\log_{30}{5}}.$$
Factor out $\log_{30}{x}$:
$$\log_2{x}+\log_3{x}+\log_5{x} = \log_{30}x\left(\frac{1}{\log_{30}{2}}+\frac{1}{\log_{30}{5}}+\frac{1}{\log_{30}{5}}\right).$$
Now use the fact that $\frac{1}{\log_b{a}} = \log_a{b}$:
$$\log_2{x}+\log_3{x}+\log_5{x} = \log_{30}{x}\left(\log_2{30}+\log_3{30}+\log_5{30}\right).$$
So now all we need to show is
$$\log_2{30}+\log_3{30}+\log_5{30}>9.$$
$\log_2{16}=4$, and since $\log_2$ is a strictly increasing function, this means $\log_2{30}>(\log_2{16}=)4$. Applying the same logic to the other two $\log$s, we get that $\log_3{30}>3$ and $\log_5{30}>2$, hence 
$$\log_2{30}+\log_3{30}+\log_5{30}>4+3+2=9$$
and we are done
A: From here since $\ln x>0$
$$\frac{\ln x}{\ln 2}+\frac{\ln x}{\ln 3}+\frac{\ln x}{\ln 5}>9\frac{\ln x}{\ln 30}\iff \frac{1}{\ln 2}+\frac{1}{\ln 3}+\frac{1}{\ln 5}>\frac{9}{\ln 30}$$
we can conclude, indeed by HM-AM inequality we have
$$\frac{3}{\frac{1}{\ln 2}+\frac{1}{\ln 3}+\frac{1}{\ln 5}}<\frac{\ln 2+\ln 3+\ln 5}{3}=\frac{\ln 30}{3}
\implies
\frac{1}{\ln 2}+\frac{1}{\ln 3}+\frac{1}{\ln 5}>\frac9{\ln 30}$$
A: $$log_2(x)+log_3(x)+log_5(x)>9log_{30}(x) \iff \frac {log(x)}{log2} + \frac {log(x)}{log3}+\frac {log(x)}{log5} >9 \frac {log(x)}{log30}$$
$$\iff \frac {1}{log2} + \frac {1}{log3}+\frac {1}{log5} >9 \frac {1}{log30}$$
$$\iff \log_2 30  +\log_3 30 + \log_5 30     > \log_2 16  +\log_3 27 + \log_5 25 =4+3+2=   9 $$
