Fie \( x,y,x \in \mathbb{R}^{\ast} \) astfel incat \( x+\frac{1}{x}=a;y+\frac{1}{y}=b;xy+\frac{1}{xy}=c \)
Demonstrati ca \( a^2+b^2+c^2=abc+4 \)
Dependenta
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Marius Mainea
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Dependenta
Last edited by Marius Mainea on Sat Jan 10, 2009 7:15 pm, edited 1 time in total.
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Andi Brojbeanu
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\( abc+4=(x+\frac{1}{x})(y+\frac{1}{y})(xy+\frac{1}{xy})+4=(xy+\frac{x}{y}+\frac{y}{x}+\frac{1}{xy})(xy+\frac{1}{xy})+4=(xy+\frac{1}{xy})(xy+\frac{1}{xy})+(xy+\frac{1}{xy})(\frac{x}{y}+\frac{y}{x})+4=(xy+\frac{1}{xy})^2+ \)
\( +(xy\cdot \frac{x}{y}+xy\cdot \frac{y}{x}+\frac{1}{xy}\cdot \frac{x}{y}+\frac{1}{xy}\cdot \frac{y}{x})+4=(xy+\frac{1}{xy})^2+(x^2+y^2+\frac{1}{y^2}+\frac{1}{x^2})+2+2=(xy+\frac{1}{xy})^2+(x^2+2\cdot x\cdot \frac{1}{x}+\frac{1}{x^2})+(y^2+2\cdot y\cdot \frac{1}{y}+\frac{1}{y^2})= \)
\( =(xy+\frac{1}{xy})^2+(x+\frac{1}{x})^2+(y+\frac{1}{y})^2=a^2+b^2+c^2 \).
\( +(xy\cdot \frac{x}{y}+xy\cdot \frac{y}{x}+\frac{1}{xy}\cdot \frac{x}{y}+\frac{1}{xy}\cdot \frac{y}{x})+4=(xy+\frac{1}{xy})^2+(x^2+y^2+\frac{1}{y^2}+\frac{1}{x^2})+2+2=(xy+\frac{1}{xy})^2+(x^2+2\cdot x\cdot \frac{1}{x}+\frac{1}{x^2})+(y^2+2\cdot y\cdot \frac{1}{y}+\frac{1}{y^2})= \)
\( =(xy+\frac{1}{xy})^2+(x+\frac{1}{x})^2+(y+\frac{1}{y})^2=a^2+b^2+c^2 \).