If $$x 0$$, $$y0$$, $$z0$$, such that $$x^2+y^2+z^2=1$$, find the minimum value of $$\dfrac{5-8xy}{z}$$ .

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

Uniteasy Admin · Accepted Answer

$$\because 2xy\leq x^2+y^2=1-z^2$$
Substitute to th original expression and use the Geometric Mean Inequality formula. 
Then, the original fraction can be converted to
$$\dfrac{5-8xy}{z}\geq \dfrac{5-1+z^2}{z} =\dfrac{4+z^2}{z} $$
$$=\dfrac{4}{z}+z $$
$$\geq 2\sqrt{\dfrac{4}{z} \cdot z}=4$$
Therefore, minimum value of the expression is equal to 4.

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