Prove the following inequality for $a,b,c, \in (1,\infty)$ , such that $abc = 10$ $$\log_a 10 + \log_b 10 + \log_c 10 \ge \sqrt{3 \log_a 10 * \log_b 10 * \log_c 10}$$
I will transform logarithms to base 10, so we have to prove that $$\frac{1}{\log_{10} a}+\frac{1}{\log_{10} b}+\frac{1}{\log_{10} c} \ge \sqrt{\frac{3}{\log_{10} a * \log_{10} b * \log_{10} c}}$$ We can use now the Cauchy inequality : $$\frac{1}{\log_{10} a}+\frac{1}{\log_{10} b}+\frac{1}{\log_{10} c} \ge \frac{(1+1+1)^2}{\log_{10} a + \log_{10} b + \log_{10} c} = \frac{9}{\log_{10} abc} = 9 $$ So we only have to prove that $$ 9 \ge \sqrt{\frac{3}{\log_{10} a * \log_{10} b * \log_{10} c}} \iff 27\log_{10} a * \log_{10} b * \log_{10} c \ge 1 $$ I got stuck here. I think we will have to continue using the inequality of means, since the numbers are positive, but I don't see where this can lead. Maybe I started wrong. What do you think ? I am here for any idea or solution you have. Thank you very much !
Apply the inequality $(x+y+z)^2 \ge 3(xy+yz+zx)$ :
$$\begin{align} \left(\frac{1}{\log_{10} a}+\frac{1}{\log_{10} b}+\frac{1}{\log_{10} c}\right)^2 &\ge 3\left(\frac{1}{\log_{10} a}\frac{1}{\log_{10} b}+\frac{1}{\log_{10} b}\frac{1}{\log_{10} c}+\frac{1}{\log_{10} c}\frac{1}{\log_{10} a} \right)\\ &= 3\cdot\frac{\log_{10} a + \log_{10} b + \log_{10} c}{\log_{10} a\cdot \log_{10} b \cdot\log_{10} c}\\&= \frac{3}{\log_{10} a\cdot \log_{10} b \cdot\log_{10} c} \end{align}$$ The equality occurs if and only if $a = b= c = \sqrt[3]{10}$.