来自倪全宝的问题
均值不等式证明若a、b、c是正数求证a^2/(b十c)十b^2/(a十c)十c^2/(a十b)>=(a+b十c)/2
均值不等式证明
若a、b、c是正数求证a^2/(b十c)十b^2/(a十c)十c^2/(a十b)>=(a+b十c)/2
1回答
2020-05-04 05:29
均值不等式证明若a、b、c是正数求证a^2/(b十c)十b^2/(a十c)十c^2/(a十b)>=(a+b十c)/2
均值不等式证明
若a、b、c是正数求证a^2/(b十c)十b^2/(a十c)十c^2/(a十b)>=(a+b十c)/2
由柯西不等式:(a^2/(b十c)十b^2/(a十c)十c^2/(a十b))*[(b+c)+(a+c)+(a+b)]
≥(a+b+c)^2
所以a^2/(b十c)十b^2/(a十c)十c^2/(a十b)>=(a+b十c)/2