Ecuatie cu parti intregi

[Marius Mainea]

Sa se afle \( x,y,z\in\mathbb{N}^* \) astfel incat \( \[\frac{x^2}{y+z}\]+\[\frac{y^2}{z+x}\]+\[\frac{z^2}{x+y}\]=\[\frac{2}{x+y+z}\] \)

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[alex2008]

\( x+y+z\ge 3\Rightarrow \left\lfloor \frac{2}{x+y+z}\right\rfloor=0 \)

Tinand cont ca \( x,y,z \) sunt numere naturale de aici rezulta ca fiecare parte intreaga este \( 0 \).

Deci \( \frac{x^2}{y+z}<1\Rightarrow x^2<y+z\Rightarrow x^2+1\le y+z \).

Analog \( y^2+1\le z+x \) si \( z^2+1\le x+y \) . Adunand obtinem \( (x-1)^2+(y-1)^2+(z-1)^2\le 0\Rightarrow x=y=z=1 \).

Parca s-a mai postat ... de catre mine.

[Claudiu Mindrila]

Cum \( x,\ y,\ z\in\mathbb{N}^{*}\Longrightarrow a+b+c\ge3\Longrightarrow\frac{2}{x+y+z}\le\frac{2}{3}<1\Longrightarrow\left[\frac{2}{x+y+z}\right]=0\ (1). \)

Dar \( \left[\frac{x^{2}}{y+z}\right],\ \left[\frac{y^{2}}{z+x}\right],\ \left[\frac{z^{2}}{x+y}\right]\ge0\Longrightarrow\sum\left[\frac{x^{2}}{y+z}\right]\ge0\ (2) \).

Acum \( \left(1\right),\ \left(2\right)\Longrightarrow\left[\frac{x^{2}}{y+z}\right]=0,\ \ \left[\frac{y^{2}}{z+x}\right]=0,\ \left[\frac{z^{2}}{x+y}\right]=0\Longrightarrow\frac{x^{2}}{y+z},\ \frac{y^{2}}{z+x},\ \frac{z^{2}}{x+y}\in\left[0,\ 1\right)\Longrightarrow\left\{ \begin{array}{c}
x^{2}<y+z\\
y^{2}<z+x\\
z^{2}<x+y\end{array}\right.\Longrightarrow\left\{ \begin{array}{c}
x^{2}+1\le y+z\\
y^{2}+1\le z+x\\
z^{2}+1\le x+y\end{array}\right. \). Insumand aceste relatii rezulta \( \sum\left(x-1\right)^{2}\le0\Longrightarrow x=y=z=1 \)

[mihai++]

Same solution.

[Claudiu Mindrila]

Nu am observat ca si Alex a postat o solutie.