Question

Cho a,b,c dương thỏa mãn: \(a^2+b^2+c^2=3\).CMR: \(a^{2019}+b^{2019}+c^{2019}\ge3\)

Answer 1

\(a^{2019}+a^{2019}+1+1+...+1\ge2019a^2\) (2017 số 1)

\(\Leftrightarrow2a^{2019}+2017\ge2019a^2\)

Tương tự: \(2b^{2019}+2017\ge2019b^2\) ; \(2c^{2019}+2017\ge2019c^2\)

Cộng vế với vế:

\(2\left(a^{2019}+b^{2019}+c^{2019}\right)+2017.3\ge2019\left(a^2+b^2+c^2\right)\)

\(\Rightarrow a^{2019}+b^{2019}+c^{2019}\ge\frac{2019\left(a^2+b^2+c^2\right)-2017.3}{2}=3\)

Dấu "=" xảy ra khi \(a=b=c=1\)