Question

Let a,b,c belong (1,infinity) and m,n belong (0,infinity). Prove that log base ((b^m)*(c^n)) a + log base ((c^m)*(a^n)) b + log base ((a^m)*(b^n)) c >= 3/(m+n)

Answer 1

If we will take that $a \geq b \geq c$, then we will have:

$$\log_{b^{m}c^{n}} a + \log_{c^{m}a^{n}} b + \log_{a^{m}b^{n}} c \geq \frac{3}{(m + n)}$$ $$\begin{array}{l} \log_{b^{m}c^{n}} a + \log_{c^{m}a^{n}} b + \log_{a^{m}b^{n}} c = \\ = \frac{1}{\log_{a} b^{m}c^{n}} + \frac{1}{\log_{b} c^{m}a^{n}} + \frac{1}{\log_{c} a^{m}b^{n}} = \\ = \frac{1}{m\log_{a} b + n\log_{a} c} + \frac{1}{m\log_{b} c + n\log_{b} a} + \frac{1}{m\log_{c} a + n\log_{c} b} \geq \\ \geq \frac{1}{m + n} + \frac{1}{m + n} + \frac{1}{m + n} = \frac{3}{m + n}. \end{array}$$

Proved.

Answer: Proved.