If 2x-3y3z-y-z-yz-x-x-3z2y-3x then prove that every ratio -xy

2x 3y3z + y = z yz x = x + 3z2y 3y Now, by use at componendo and devendeo we get ∴xy = xy = xy Hence proved ...

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Componendo and Dividendo

If 2x-3y3z+...

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If $\frac{2 x - 3 y}{3 z + y} = \frac{z - y}{z - x} = \frac{x + 3 z}{2 y - 3 x}$ then prove that every ratio $= \frac{x}{y}$

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x^{3} y^{3} + y^{3} z^{3} + z^{3} x^{3} \geq 3 x^{2} y^{2} z^{2}

3 x (x - y - z) + 3 y (x - y + z) + 3 z (x + y - z)

3 (x + y) (x - y) + 3 z (2 y + z)

\frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10 \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10 \frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13

\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2; x, y, z \neq 0

x + y + z = x y z

\frac{3 x - x^{3}}{1 - 3 x^{2}} + \frac{3 y - y^{3}}{1 - 3 y^{2}} + \frac{3 z - z^{3}}{1 - 3 z^{2}} = \frac{3 x - x^{3}}{1 - 3 x^{2}} \cdot \frac{3 y - y^{3}}{1 - 3 y^{2}} \cdot \frac{3 z - z^{3}}{1 - 3 z^{2}}

∣ ∣ ∣ ∣ \begin{matrix} x & x + y & x + y + z 2 x & 3 x + 2 y & 4 x + 3 y + 2 z 3 x & 6 x + 3 y & 10 x + 6 y + 3 z \end{matrix} ∣ ∣ ∣ ∣

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