\(x^2+2xy+2y^2+y+\frac{1}{2}\)
\(=x^2+2xy+y^2+y^2+y+\frac{1}{2}\)
\(=\left(x+y\right)^2+y^2+y+\frac{1}{2}\)
Có: \(\left(x+y\right)^2\ge0\)
\(y^2\ge y\ge0\Rightarrow y^2+y\ge0\)
\(\frac{1}{2}>0\)
\(\Rightarrow x^2+2xy+2y^2+y+\frac{1}{2}>0\) với mọi x
xét vế trái: \(x^2+2xy+2y^2+y+\frac{1}{2}\) =\(x^2+2xy+y^2+y^2+y+\frac{1}{2}\)
= \(\left(x^2+2xy+y^2\right)+\left(y^2+y+\frac{1}{2}\right)\)
= \(\left(x+y\right)^2+\left(y^2+2.\frac{1}{2}.y+\frac{1}{4}-\frac{1}{4}+\frac{1}{2}\right)\)
= \(\left(x+y\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{1}{4}\)
vì \(\left(x+y\right)^2>=0\) và \(\left(y+\frac{1}{2}\right)^2>=0\) => \(\left(x+y\right)^2+\left(y+\frac{1}{2}\right)^2>=0\)
mà 1/4 >0 => \(\left(x+y\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{1}{4}>0\)