# Need some hints for proving a logarithmic inequality.

logaxlogabx+logbxlogbcx+logcxlogacx≥6\frac{\log_ax}{\log_{ab}x} +
\frac{\log_b{x}}{\log_{bc}x} +
\frac{\log_cx}{\log_{ac}x} \ge 6

Did as you suggested and got this, im stuck again:

logab+logbc+logca≥3\log_ab + \log_bc + \log_ca \ge 3

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– Redundant Aunt
2 days ago

logaxlogabx=1/logxa1/logx(ab)=logx(ab)logxa=loga(ab)=logaa+logab=1+logab. \frac{\log_a x}{\log_{ab} x} = \frac{1/\log_x a}{1/\log_x(ab)} = \frac{\log_x (ab)}{\log_x a} = \log_a(ab) = \log_a a + \log_a b = 1 + \log_a b. Therefore logaxlogabx+logbxlogbcx+logcxlogacx=(1+logab)+(1+logbc)+(1+logca) \frac{\log_ax}{\log_{ab}x} + \frac{\log_b{x}}{\log_{bc}x} + \frac{\log_cx}{\log_{ac}x} = (1+ \log_a b) + (1+ \log_b c) + (1+ \log_c a) =3+(logab+logbc+logca)=3+logbloga+logclogb+logalogc=3+uv+wu+vw= 3 + (\log_a b + \log_b c + \log_c a) = 3 + \frac{\log b}{\log a} + \frac{\log c}{\log b} + \frac{\log a}{\log c} = 3 + \frac u v + \frac w u + \frac v w
– Michael Hardy
2 days ago

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4

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Let m=logxa,n=logxb,p=logxc⟹LHS=3+nm+pn+mp≥3+3=6m = \log_x a, n = \log_x b, p = \log_x c\implies LHS = 3 + \dfrac{n}{m}+\dfrac{p}{n}+ \dfrac{m}{p} \ge 3+3 = 6 by applying AM-GM inequality.

Consider making use of these three identities:

logax=logbxlogba\log_{a}{x}=\frac{\log_{b}{x}}{\log_{b}{a}}
logqa+logqb=logqab\log_{q}{a}+\log_{q}{b}=\log_{q}{ab}
logbab=1+logba\log_{b}{ab}=1+\log_{b}{a}

If you write logaxlogabx=lnxlnalnxlnab=lnablna=1+lnblna\frac{\log_a x}{\log_{ab} x}=\frac{\frac{\ln x}{\ln a}}{\frac{\ln x}{\ln ab}} = \frac{\ln ab}{\ln a} = 1+\frac{\ln b}{\ln a}, and something similar with the other fractions, you can the use the familiar inequality :
xy+yz+zx≥3\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge3
which can be derived easily from the inequality between arithmetic and geometric means.

logaxlogabx=1/logxa1/logx(ab)=logx(ab)logxa=loga(ab)=logaa+logab=1+logab.
\frac{\log_a x}{\log_{ab} x} = \frac{1/\log_x a}{1/\log_x(ab)} = \frac{\log_x (ab)}{\log_x a} = \log_a(ab) = \log_a a + \log_a b = 1 + \log_a b.

Therefore
logaxlogabx+logbxlogbcx+logcxlogacx=(1+logab)+(1+logbc)+(1+logca)=3+(logab+logbc+logca)=3+logbloga+logclogb+logalogc=3+uv+wu+vw\begin{align}
& \frac{\log_ax}{\log_{ab}x} + \frac{\log_b{x}}{\log_{bc}x} + \frac{\log_cx}{\log_{ac}x} = (1+ \log_a b) + (1+ \log_b c) + (1+ \log_c a) \\[10pt]
= {} & 3 + (\log_a b + \log_b c + \log_c a) = 3 + \frac{\log b}{\log a} + \frac{\log c}{\log b} + \frac{\log a}{\log c} = 3 + \frac u v + \frac w u + \frac v w
\end{align}