Ta có :
\(y^2=xz\Leftrightarrow\dfrac{x}{y}=\dfrac{y}{z}\left(1\right)\)
\(z^2=yt\Leftrightarrow\dfrac{x}{y}=\dfrac{t}{x}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{t}{x}\)
\(\Leftrightarrow\dfrac{x^3}{y^3}=\dfrac{y^3}{z^3}=\dfrac{t^3}{x^3}\)
Áp dụng t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{x^3}{y^3}=\dfrac{y^3}{z^3}=\dfrac{t^3}{x^3}=\dfrac{x^3+y^3+t^3}{y^3+z^3+x^3}\)
\(\Leftrightarrow\dfrac{x^3}{t^3}=\dfrac{y^3+z^3+x^3}{y^3+z^3+x^3}\left(đpcm\right)\)