Fie x, y, z >0 astfel incat x+y+z=xyz. Demonstrati ca:
\( \frac{x+y}{1+z^2}+\frac{y+z}{1+x^2}+\frac{z+x}{1+y^2}\ge \frac{27}{2xyz} \)
Marius Mainea
Inegalitate conditionata de x+y+z=xyz (own)
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Marius Mainea
- Gauss
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Marius Damian
- Arhimede
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Notam \( \frac{1}{xy}=a>0 \), \( \frac{1}{yz}=b>0 \), \( \frac{1}{zx}=c>0 \). Inegalitatea din enunt se scrie echivalent \( \sum{\frac{b+c}{a(a+b)(a+c)}}\geq\frac{27}{2} \), conditionata de \( a+b+c=1 \).
Folosind inegalitatile Cauchy-Schwarz si AM-GM, avem
\( \sum{\frac{b+c}{a(a+b)(a+c)}}=\frac{1}{(a+b)(b+c)(c+a)}\cdot\sum{\frac{(b+c)^2}{a}}\geq\frac{1}{(\frac{a+b+b+c+c+a}{3})^3}\cdot\frac{(a+b+b+c+c+a)^2}{a+b+c}=\frac{27}{2} \) si inegalitatea este demonstrata.
Folosind inegalitatile Cauchy-Schwarz si AM-GM, avem
\( \sum{\frac{b+c}{a(a+b)(a+c)}}=\frac{1}{(a+b)(b+c)(c+a)}\cdot\sum{\frac{(b+c)^2}{a}}\geq\frac{1}{(\frac{a+b+b+c+c+a}{3})^3}\cdot\frac{(a+b+b+c+c+a)^2}{a+b+c}=\frac{27}{2} \) si inegalitatea este demonstrata.