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Prove inequality $$\log_ab+\log_bc+\log_ca\geq1+\log_{ab}bc+\log_{bc}ab$$ for $a>1,b>1,c>1.$

Inequality is interesting because of asymmetry and inhomogeneity and I think the solution might interest someone.

We noted $x=\lg a,y=\lg b, z=\lg c $ and wrote inequality in the form

$$\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\geq 1+\frac{y+z}{x+y}+\frac{x+y}{y+z}$$ for $x>0, y>0, z>0.$

We denote $$ \frac{y}{x}=A,\frac{z}{y}=B, \frac{x}{z}=C$$ with $ABC=1$ and we get $$A+B+C\geq \frac{A+AB}{1+A}+\frac{1+A}{A+AB}+1.$$ For $C=\frac{1}{AB}$ is obtained $$\begin{align} A^3B^2+A^3B+AB^3-A^2B^2-2A^2B-2AB+A+1 &\geq0 \\ \implies\quad\quad\quad (A+1)(AB-1)^2+AB(A-B)^2 &\geq0. \end{align}$$

This is a solution. Has anyone another idea?

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  • $\begingroup$ Nice solution! A minor suggestion: use \iff for <=>. $\endgroup$ – Kim Jong Un Nov 30 '14 at 21:14
  • $\begingroup$ Although this is from one year ago, I just asked a similar question here. $\endgroup$ – Colescu Apr 25 '16 at 11:38
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We can use C-S.

Indeed, $\frac{x}{z}+\frac{y}{x}+\frac{z}{y}-1-\frac{x+y}{y+z}-\frac{y+z}{x+y}=\frac{x^2}{xz}+\frac{y^2}{yx}+\frac{z^2}{yz}+\frac{y^2}{y^2}-\left(\frac{x+y}{y+z}+\frac{y+z}{x+y}+2\right)\geq$

$\geq\frac{(x+y+z+y)^2}{xz+yx+yz+y^2}-\left(\frac{x+y}{y+z}+\frac{y+z}{x+y}+2\right)=0$

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  • $\begingroup$ Very ingenious. Thank you! $\endgroup$ – medicu Nov 30 '16 at 12:33

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