Question

Cho a, b, c là các số thực. Chứng minh:

a²⁰¹⁰ + b²⁰¹⁰ + c²⁰¹⁰> a¹⁰⁰⁵b¹⁰⁰⁵ + a¹⁰⁰⁵c¹⁰⁰⁵ + c¹⁰⁰⁵b¹⁰⁰⁵

Answer 1

Ta có \(\left(a^{1005}-b^{1005}\right)^2+\left(b^{1005}-c^{1005}\right)^2+\left(c^{1005}-a^{1005}\right)^2>0\Leftrightarrow a^{2010}-2a^{1005}b^{1005}+b^{2010}+b^{2010}-2b^{1005}c^{1005}+c^{2010}+c^{2010}-2a^{1005}c^{1005}+a^{1005}>0\Leftrightarrow2\left(a^{2010}+b^{2010}+c^{2010}\right)-2\left(a^{1005}b^{1005}+a^{1005}c^{1005}+c^{1005}b^{1005}\right)>0\Leftrightarrow a^{2010}+b^{2010}+c^{2010}>a^{1005}b^{1005}+a^{1005}c^{1005}+c^{1005}b^{1005}\)(đpcm)

Answer 2

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