matematikk.net

Gitt at $xy+yz+zx=1$, vis da at

$ \displaystyle \hspace{1cm}
\frac{x}{1+x^2} + \frac{y}{1+y^2} + \frac{z}{1+z^2}
=
\frac{ 2 }{ \sqrt{ (1+x^2)(1+y^2)(1+z^2) }\:}
$

Godt mulig det finnes finere måter å gjøre dette på!

Vi har $ 1 = 1^2 = (xy + xz + yz)^2 = x^2 y^2 + x^2 z^2 + y^2 z^2 + 2(x^2 yz + x y^2 z + xy z^2 ) $
slik at $ 2(x^2 yz + x y^2 z + xy z^2 ) = 1 - x^2 y^2 - x^2 z^2 - y^2 z^2 $.

Vi har også $$ \frac{x}{1 + x^2} + \frac{y}{1 + y^2} + \frac{z}{1 + z^2} = \frac{x(1+y^2)(1 + z^2) + y(1+x^2)(1+z^2) + z(1+x^2)(1+y^2)}{(1+x^2)(1+y^2)(1+z^2)} $$

Nå kan vi skrive $$ x(1+y^2)(1 + z^2) + y(1+x^2)(1+z^2) + z(1+x^2)(1+y^2) = \sqrt{(x(1+y^2)(1 + z^2) + y(1+x^2)(1+z^2) + z(1+x^2)(1+y^2))^2} $$
$$ = \sqrt{x^2 (1+y^2)^2 (1 + z^2)^2 + y^2 (1 + x^2)^2 (1 + z^2)^2 + z^2 (1+x^2)^2 (1 + y^2)^2 + 2 xy (1+x^2)(1+y^2)(1+z^2)^2 + 2 xz (1 + x^2)(1+y^2)^2(1+z^2) + 2yz(1+x^2)^2(1+y^2)(1+z^2)} $$
$$ = \sqrt{(1+x^2)(1+y^2)(1+z^2) \bigg( \frac{x^2(1+y^2)(1+z^2)}{1+x^2} + \frac{y^2(1+x^2)(1+z^2)}{1+y^2} + \frac{z^2(1+x^2)(1+y^2)}{1+z^2} + 2(xy + xz + yz) + 2(x^2yz + xy^2 z + xyz^2) \bigg) } $$
$$ = \sqrt{(1+x^2)(1+y^2)(1+z^2) \bigg( \frac{1 + x^2 - 1 + x^2 y^2 + x^2 z^2 + x^2 y^2 z^2}{1+x^2} + \frac{1 + y^2 - 1 + x^2 y^2 + y^2 z^2 + x^2 y^2 z^2}{1+y^2} + \frac{1 + z^2 - 1 + x^2 z^2 + y^2 z^2 + x^2 y^2 z^2}{1+z^2} + 2 + 1 - x^2 y^2 - x^2 z^2 - y^2 z^2 \bigg) } $$
$$ = \sqrt{(1+x^2)(1+y^2)(1+z^2) \bigg( 1 + \frac{ - 1 + x^2 y^2 + x^2 z^2 + x^2 y^2 z^2}{1+x^2} + 1 + \frac{ - 1 + x^2 y^2 + y^2 z^2 + x^2 y^2 z^2}{1+y^2} + 1 + \frac{ - 1 + x^2 z^2 + y^2 z^2 + x^2 y^2 z^2}{1+z^2} + 3 - \frac{x^2 y^2 - x^2 y^2 z^2}{1 + z^2} - \frac{x^2 z^2 - x^2 y^2 z^2}{1 + y^2} - \frac{y^2 z^2 + x^2 y^2 z^2}{1 + x^2} \bigg) } $$
$$ = \sqrt{(1+x^2)(1+y^2)(1+z^2) \bigg(6 + \frac{-( y^2 z^2 + 2(x^2 yz + xy^2 z + xy z^2)) + x^2 y^2 z^2 - (y^2 z^2 + x^2 y^2 z^2)}{1 + x^2} + \frac{-(x^2 z^2 + 2(x^2 yz + xy^2 z + xy z^2)) + x^2 y^2 z^2 - (x^2 z^2 - x^2 y^2 z^2)}{1 + y^2} + \frac{-(x^2 y^2 + 2(x^2 yz + xy^2 z + xy z^2)) + x^2 y^2 z^2 - (x^2 y^2 - x^2 y^2 z^2)}{1 + z^2} \bigg) } $$
$$ = \sqrt{(1+x^2)(1+y^2)(1+z^2) \bigg(6 - 2yz \frac{xy + xz + yz + x^2}{1 + x^2} - 2xz \frac{xy + xz + yz + y^2}{1+ y^2} - 2xy \frac{xy + xz + yz + x^2}{1+z^2} \bigg) } $$
$$ = \sqrt{4(1+x^2)(1+y^2)(1+z^2)} = 2\sqrt{(1+x^2)(1+y^2)(1+z^2)} $$

Da har vi $$ \frac{x}{1 + x^2} + \frac{y}{1 + y^2} + \frac{z}{1 + z^2} = \frac{2\sqrt{(1+x^2)(1+y^2)(1+z^2)}}{(1+x^2)(1+y^2)(1+z^2)} = \frac{2}{\sqrt{(1+x^2)(1+y^2)(1+z^2)}} $$

Alternativt la $x = \tan A$, $y = \tan B$ og $z = \tan C$

Ironisk nok var en slik substitusjon det første som falt meg inn. Dessverre så jeg ikke umiddelbart hvordan det gjorde problemet nevneverdig mye enklere, så jeg gikk heller for den grusomt stygge veien i steden.