Fie \( a,b,c >0 \) cu \( abc=1 \). Demonstrati ca urmatoarea inegalitate are loc:
\( \displaystyle \sum_{cyc}\frac{1}{(a+b)b} \geq \frac{3}{2} \)
Zhautykov Olympiad 2008
Inegalitate conditionata, trei variabile, ciclica
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Vlad Matei
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Cezar Lupu
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Intr-adevar, cum produsul numerelor \( a, b, c \) este \( 1 \), putem lua \( a=\frac{x}{y}, b=\frac{y}{z} \) si \( c=\frac{z}{x} \). Astfel inegalitatea devine
\( \sum_{cyc}\frac{z^{2}}{zx+y^{2}}\geq\frac{3}{2} \).
Insa, din inegalitatea \( z^{2}+x^{2}\geq 2zx \), vom obtine ca
\( \sum_{cyc}\frac{z^{2}}{zx+y^{2}}\geq\sum_{cyc}\frac{2z^{2}}{z^{2}+x^{2}+2y^{2}} \).
Acum mai ramane sa demonstram ca
\( \sum_{cyc}\frac{2z^{2}}{z^{2}+x^{2}+2y^{2}}\geq\frac{3}{2} \)
care se face folosind din nou inegalitatea Cauchy-Schwarz.
\( \sum_{cyc}\frac{z^{2}}{zx+y^{2}}\geq\frac{3}{2} \).
Insa, din inegalitatea \( z^{2}+x^{2}\geq 2zx \), vom obtine ca
\( \sum_{cyc}\frac{z^{2}}{zx+y^{2}}\geq\sum_{cyc}\frac{2z^{2}}{z^{2}+x^{2}+2y^{2}} \).
Acum mai ramane sa demonstram ca
\( \sum_{cyc}\frac{2z^{2}}{z^{2}+x^{2}+2y^{2}}\geq\frac{3}{2} \)
care se face folosind din nou inegalitatea Cauchy-Schwarz.
Last edited by Cezar Lupu on Mon Jan 28, 2008 7:55 pm, edited 1 time in total.
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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Cezar Lupu
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Solutia 2. (pe ideea din concurs a lui Flavian Georgescu)
Fara a leza generalitatea problemei, sa presupunem ca \( a\leq b\leq c \). Din inegalitatea rearanjamentelor avem ca
\( A=\sum_{cyc}\frac{1}{b(a+b)}\geq\sum_{cyc}\frac{1}{c(a+b)}=B \).
Prin urmare \( 2A\geq A+B=\sum_{cyc}\frac{1}{b(a+b)}+\sum_{cyc}\frac{1}{c(a+b)}=\sum_{cyc}\frac{ac}{a+b}+\sum_{cyc}\frac{ab}{a+b}=\sum_{cyc}\frac{a(b+c)}{a+b} \).
Acum, aplicand inegalitatea mediilor, avem ca \( \sum_{cyc}\frac{a(b+c)}{a+b}\geq 3 \), de unde \( A\geq\frac{3}{2} \).
Fara a leza generalitatea problemei, sa presupunem ca \( a\leq b\leq c \). Din inegalitatea rearanjamentelor avem ca
\( A=\sum_{cyc}\frac{1}{b(a+b)}\geq\sum_{cyc}\frac{1}{c(a+b)}=B \).
Prin urmare \( 2A\geq A+B=\sum_{cyc}\frac{1}{b(a+b)}+\sum_{cyc}\frac{1}{c(a+b)}=\sum_{cyc}\frac{ac}{a+b}+\sum_{cyc}\frac{ab}{a+b}=\sum_{cyc}\frac{a(b+c)}{a+b} \).
Acum, aplicand inegalitatea mediilor, avem ca \( \sum_{cyc}\frac{a(b+c)}{a+b}\geq 3 \), de unde \( A\geq\frac{3}{2} \).
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.