Sã se arate cã dacã x, y, z sunt numere reale nenule astfel încât x + y + z = 0, atunci
\( \frac{x^2 + y^2}{x+y} + \frac{y^2 + z^2}{y+z} + \frac{z^2 + x^2}{z+x} = \frac{x^3}{yz} + \frac{y^3}{zx} + \frac{z^3}{xy} \).
O problema draguta
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O problema draguta
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Elev la Colegiul Naţional Ienăchiţă-Văcărescu, Târgovişte,
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Elev la Colegiul Naţional Ienăchiţă-Văcărescu, Târgovişte,
Clasa a 8 a
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Mateescu Constantin
- Newton
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\( LHS=\sum\ \frac{(y+z)^2-2yz}{y+z}=\sum\ \left(-x+\frac{2yz}x\right)=\sum\ \frac{2yz}{x}=\frac{1}{xyz}\ \cdot\ \sum\ 2y^2z^2=\frac{1}{xyz}\ \cdot\ \sum\ x^4=RHS \)
Deci ramane sa aratam ca : \( x^4+y^4+z^4=2(x^2y^2+y^2z^2+z^2x^2)\ \Longleftrightarrow\ (x^2-y^2+z^2)^2-(2zx)^2=0 \)
\( \Longleftrightarrow\ (x^2-y^2+z^2-2zx)(x^2-y^2+z^2+2zx)=0\ \Longleftrightarrow\ [(x-z)^2-y^2][(x+z)^2-y^2]=0\ \Longleftrightarrow\ \)
\( \Longleftrightarrow\ (x-y-z) (x+y-z) (x-y+z) (x+y+z)=0 \), ceea ce este adevarat intrucat \( x+y+z=0 \) .
Deci ramane sa aratam ca : \( x^4+y^4+z^4=2(x^2y^2+y^2z^2+z^2x^2)\ \Longleftrightarrow\ (x^2-y^2+z^2)^2-(2zx)^2=0 \)
\( \Longleftrightarrow\ (x^2-y^2+z^2-2zx)(x^2-y^2+z^2+2zx)=0\ \Longleftrightarrow\ [(x-z)^2-y^2][(x+z)^2-y^2]=0\ \Longleftrightarrow\ \)
\( \Longleftrightarrow\ (x-y-z) (x+y-z) (x-y+z) (x+y+z)=0 \), ceea ce este adevarat intrucat \( x+y+z=0 \) .