Question

1. Giải hệ \(\left\{{}\begin{matrix}x+y+z=2\\2xy-z^2=4\end{matrix}\right.\)

2. Cho x, y, z>0 thỏa mãn: x+y+z=2

Tìm giá trị nhỏ nhất của P\(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)

Answer 1

\(\left\{{}\begin{matrix}x=2-y-z\\z^2-2xy+4=0\end{matrix}\right.\) \(\Rightarrow z^2-2y\left(2-y-z\right)+4=0\)

\(\Rightarrow z^2-4y+2y^2+2yz+4=0\)

\(\Rightarrow z^2+2yz+y^2+y^2-4y+4=0\)

\(\Rightarrow\left(z+y\right)^2+\left(y-2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}z+y=0\\y-2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=2\\z=-2\\x=2\end{matrix}\right.\)

b/ Áp dụng BĐT Cauchy-Schwarz:

\(P=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}=1\)

\(\Rightarrow P_{min}=1\) khi \(x=y=z=\frac{2}{3}\)