若x,y,z均为正实数,且x^2+y^2+z^2=1,则S=(z+1)^2/2xyz的最小值是
人气:136 ℃ 时间:2020-05-21 21:40:40
解答
2xy ≤ x^2+y^2 = 1 - z^2,仅当x=y时成立
∴S = (z + 1)^2 / 2xyz
≥ (z + 1)^2 / z(1 - z^2)
= (z + 1) / z(1 - z)
= - 1 / [(z+1) - 3 + 2/(z+1)]
由于(z+1) + 2/(z+1) - 3 ≥ 2√2 - 3,等号当z+1 = 2/(z+1),亦即z = √2-1时成立.
所以 - 1 / [(z+1) - 3 + 2/(z+1)] ≥ 1/(3 - 2√2) = 3 + 2√2,
S的最小值为3 + 2√2,当z = √2-1时成立.
不难求出,此时的x = y = √(√2 - 1).
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