给定正整数k,当x^k+y^k+z^k=1时,求x^(k+1)+y^(k+1)+z^(k+1)最小值0分

给定正整数k,当x^k+y^k+z^k=1时,求x^(k+1)+y^(k+1)+z^(k+1)最小值0分
给定正整数k,当x^k+y^k+z^k=1时,求x^(k+1)+y^(k+1)+z^(k+1)最小值0分

给定正整数k,当x^k+y^k+z^k=1时,求x^(k+1)+y^(k+1)+z^(k+1)最小值0分
由幂平均不等式得
[(x^(k+1)+y^(k+1)+z^(k+1))/3]^[1/(k+1)]
≥[(x^k+y^k+z^k)/3]^(1/k)
=(1/3)^(1/k),
故x^(k+1)+y^(k+1)+z^(k+1)
≥3(1/3)^[(k+1)/k]
=3^(-1/k).
当x=y=z=3^(-1/k)时等号成立.
故所求最小值为3^(-1/k).

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