Inegalitate cu logaritmi

[BogdanCNFB]

Daca \( \alpha\in (0,2] \) si daca \( a,b,c\in (0,1) \) sau \( a,b,c\in(0,\propto) \) atunci sa se arate ca:
\( \frac{1}{\alpha+\log_a{b}}+\frac{1}{\alpha+\log_b{c}}+\frac{1}{\alpha+\log_c{a}}\le \frac{2}{\alpha} \)

Dumitru Acu, Sibiu

[Marius Mainea]

Presupunem a,b,c>1. Notam \( x=\ln a \), \( y=\ln b \), \( z=\ln c \). Avem x,y,z>0 si inegalitatea devine

\( \sum {\frac{1}{\alpha+\frac{y}{x}}}\leq \frac{2}{\alpha} \)
sau
\( \sum {\frac{y}{\alpha x+y}\geq 1 \)
care este adevarata deoarece \( \alpha\leq2 \) si atunci \( \sum {\frac{y}{\alpha x+y}}\geq\sum {\frac{y}{2x+y}}=\sum {\frac{y^2}{2xy+y^2}}\geq\frac{(x+y+z)^2}{(x+y+z)^2}=1 \).

Analog daca a,b,c<1 logaritmam intr-o baza subunitara.

[BogdanCNFB]

Zi-mi si mie te rog cum treci de la \( \sum\frac{1}{\alpha+\frac{y}{x}}\leq\frac{2}{\alpha} \) la \( \sum\frac{y}{\alpha x+y}\geq 1 \).

[Marius Mainea]

Zi-mi si mie te rog cum treci de la \( \sum\frac{1}{\alpha+\frac{y}{x}}\leq\frac{2}{\alpha} \) la \( \sum\frac{y}{\alpha x+y}\geq 1 \)

\( \frac{x}{\alpha x+y}=\frac{1}{\alpha}-\frac{y}{\alpha(\alpha x+y)} \)