If x > 0, y>0, z>0, such that x^2+y^2+z^2=1, find the minimum value of \dfrac{5-8xy}{z} .

Collected in the board: Inequality

Steven Zheng posted 20 minutes ago


\because 2xy\leq x^2+y^2=1-z^2

Substitute to th original expression and use the Geometric Mean Inequality formula.

Then, the original fraction can be converted to

\dfrac{5-8xy}{z}\geq \dfrac{5-1+z^2}{z} =\dfrac{4+z^2}{z}


\geq 2\sqrt{\dfrac{4}{z} \cdot z}=4

Therefore, minimum value of the expression is equal to 4.

Steven Zheng posted 14 minutes ago

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