-Dấu lớn hơn hay lớn hơn hoặc bằng vậy bạn?

xl ạ ≤

Với mọi x;y;z ta luôn có:

\(\left(x+y-1\right)^2+\left(z-\dfrac{1}{2}\right)^2\ge0\)

\(\Leftrightarrow x^2+y^2+2xy-2x-2y+1+z^2-z+\dfrac{1}{4}\ge0\)

\(\Leftrightarrow x^2+y^2+z^2+\dfrac{5}{4}+2xy-2x-2y-z\ge0\)

\(\Leftrightarrow2+2xy-2x-2y\ge z\)

\(\Leftrightarrow2\left(1-x\right)\left(1-y\right)\ge z\)

Dấu "=" xảy ra khi và chỉ khi \(x=y=z=\dfrac{1}{2}\)

\(\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)

\(\Rightarrow2\ge3x^2+2y^2+2z^2+y^2+z^2\) 

\(\Leftrightarrow2\ge3\left(x^2+y^2+z^2\right)\)

Có: \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\le2\)

\(\Rightarrow\)\(A^2\le2\) \(\Leftrightarrow A\in\left[-\sqrt{2};\sqrt{2}\right]\)

minA=-1\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y+z=-\sqrt{2}\\x=y=z\end{matrix}\right.\)  \(\Rightarrow x=y=z=-\dfrac{\sqrt{2}}{3}\)

maxA=1\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=\sqrt{2}\\x=y=z\end{matrix}\right.\) \(\Rightarrow x=y=z=\dfrac{\sqrt{2}}{3}\)

/\(2020\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{y^2+z^2}+\dfrac{1}{x^2+y^2}\right)ápdụngBDT\)

\(\dfrac{1}{x^2+y^2}+\dfrac{1}{y^2+z^2}+\dfrac{1}{x^2+z^2}\ge\dfrac{9}{2\left(x^2+y^2+z^2\right)}=\dfrac{9}{2\cdot2020}\)

\(ápdụngBĐTcosi\)

\(x^3+y^3+z^3\ge3xyz\)

\(\)=> VP\(\ge\) 9/2

Đáp án B

Đề lỗi công thức rồi. Bạn xem lại.