Question

[Nguyễn Hoàng]

Cho các số thực a;b;c thỏa mãn:

\(\left|a\right|;\left|b\right|;\left|c\right|\le1\) và a+b+c=0

CMR \(a^{2018}+b^{2020}+c^{2022}\le2\)

Answer 1

[Nguyễn Việt Lâm]

Do \(\left|a\right|;\left|b\right|;\left|c\right|\le1\Rightarrow a^{2018}+b^{2020}+c^{2022}\le a^2+b^2+c^2\)

Đặt \(\left(a;b;c\right)=\left(x-1;y-1;z-1\right)\Rightarrow\left[{}\begin{matrix}0\le x;y;z\le2\\x+y+z=3\end{matrix}\right.\)

Ta cần chứng minh: \(\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2\le2\)

\(\Leftrightarrow x^2+y^2+z^2-2\left(x+y+z\right)+3\le2\)

\(\Leftrightarrow x^2+y^2+z^2\le5\)

Thật vậy, do \(0\le x;y;z\le2\)

\(\Rightarrow\left(2-x\right)\left(2-y\right)\left(2-z\right)\ge0\)

\(\Leftrightarrow8-4\left(x+y+z\right)+2\left(xy+yz+zx\right)-xyz\ge0\)

\(\Leftrightarrow2\left(xy+yz+zx\right)\ge xyz+4\ge4\)

\(\Leftrightarrow\left(x+y+z\right)^2-\left(x^2+y^2+z^2\right)\ge4\)

\(\Leftrightarrow x^2+y^2+z^2\le5\) (đpcm)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(0;1;2\right)\) và hoán vị

Hay \(\left(a;b;c\right)=\left(-1;0;1\right)\) và hoán vị