Daca \( 0<a\leq c\leq b<1 \), atunci
\( \log_{ab}{\frac{a}{b}}+\log_{bc}{\frac{b}{c}}+\log_{ca}{\frac{c}{a}\leq 0 \).
Inegalitate cu logaritmi
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Marius Mainea
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\( \frac{\lg{\frac{a}{b}}}{\lg{ab}}+\frac{\lg{\frac{b}{c}}}{\lg{bc}}+\frac{\lg{\frac{c}{a}}}{\lg{ca}}\leq 0 \)
\( \fra{\lg{a}-\lg{b}}{\lg{a}+lg{b}}+\fra{\lg{b}-\lg{c}}{\lg{b}+lg{c}}+\fra{\lg{c}-\lg{a}}{\lg{c}+lg{a}}\leq0 \)
Notam \( \lg{a}=x,\lg{b}=y,\lg{c}=z \)
\( \fra{x-y}{x+y}+\frac{y-z}{y+z}+\frac{z-x}{z+x}\leq0 \)
\( \frac{x}{x+y}-\frac{y}{x+y}+\frac{y}{y+z}-\frac{z}{y+z}+\frac{z}{x+z}-\frac{x}{x+z}\leq0 \)
\( (\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{x+z})-(\frac{y}{x+y}+\frac{z}{y+z}+\frac{x}{x+z})\leq0 \)
\( \sum{\frac{x^2}{x^2+yx}\geq\frac{(x+y+z)^2}{x^2+yx+zy+xz+y^2+z^2 \)
\( \sum{\frac{x^2}{x^2+zx}\geq\frac{(x+y+z)^2}{x^2+yx+zy+xz+y^2+z^2 \)
\(
=>LHS=0 \)
\( \fra{\lg{a}-\lg{b}}{\lg{a}+lg{b}}+\fra{\lg{b}-\lg{c}}{\lg{b}+lg{c}}+\fra{\lg{c}-\lg{a}}{\lg{c}+lg{a}}\leq0 \)
Notam \( \lg{a}=x,\lg{b}=y,\lg{c}=z \)
\( \fra{x-y}{x+y}+\frac{y-z}{y+z}+\frac{z-x}{z+x}\leq0 \)
\( \frac{x}{x+y}-\frac{y}{x+y}+\frac{y}{y+z}-\frac{z}{y+z}+\frac{z}{x+z}-\frac{x}{x+z}\leq0 \)
\( (\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{x+z})-(\frac{y}{x+y}+\frac{z}{y+z}+\frac{x}{x+z})\leq0 \)
\( \sum{\frac{x^2}{x^2+yx}\geq\frac{(x+y+z)^2}{x^2+yx+zy+xz+y^2+z^2 \)
\( \sum{\frac{x^2}{x^2+zx}\geq\frac{(x+y+z)^2}{x^2+yx+zy+xz+y^2+z^2 \)
\(
=>LHS=0 \)