\(x^2+2xy+2y^2+y+\frac{1}{2}\)

\(=x^2+2xy+y^2+y^2+y+\frac{1}{2}\)

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

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    =>     \(\left(x+y\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{1}{4}>0\)

a. Ta có : \(4x^2-6x+9=4x^2-6x+\dfrac{9}{4}+\dfrac{27}{4}\)

\(=\left[\left(2x\right)^2-6x+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{27}{4}\)

\(=\left(2x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\)

Vì \(\left(2x-\dfrac{3}{2}\right)^2\ge0\forall x\)

nên \(\left(2x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\forall x\)

b.Ta có : \(x^2+2y^2-2xy+y+1=\left(x^2+y^2-2xy\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\)

\(=\left(x-y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Vì \(\left(x-y\right)^2\ge0\forall x;y\)

\(\left(y+\dfrac{1}{2}\right)^2\ge0\forall y\)

nên \(\left(x-y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}>0\forall x;y\)

\(x^2+2y^2-2xy+2x-4y+2=0\)

\(\Rightarrow x^2-2xy+y^2+2\left(x-y\right)+1+y^2-2y+1=0\)

\(\Rightarrow\left(x-y\right)^2+2\left(x-y\right)+1+\left(y-1\right)^2=0\)

\(\Rightarrow\left(x-y+1\right)^2+\left(y-1\right)^2=0\)

=>................

\(x^2+2y^2-2xy+2x-4y+3\)

\(=x^2-2x\left(y-1\right)+\left(y-1\right)^2-\left(y-1\right)^2+2y^2-4y+3\)

\(=\left(x-y+1\right)^2-y^2+2y+1+2y^2-4y+3\)

\(=\left(x-y+1\right)^2+y^2-2y+4\)

\(=\left(x-y+1\right)^2+\left(y-1\right)^2+3>0\forall x;y\)

Áp dụng BĐT \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\):

\(VT=\sqrt{\frac{x^2+\left(2y\right)^2}{2}}+\sqrt{\frac{\left(\frac{x}{2}-y\right)^2+3\left(\frac{x}{2}+y\right)^2}{3}}\)

\(VT\ge\sqrt{\frac{\left(x+2y\right)^2}{4}}+\sqrt{\frac{3\left(\frac{x}{2}+y\right)^2}{3}}\)

\(VT\ge\left|\frac{x+2y}{2}\right|+\left|\frac{x+2y}{2}\right|=\left|x+2y\right|\ge x+2y\) (đpcm)

Dấu "=" xảy ra khi \(x=2y\ge0\)

Đẹp Trai Không Bao Giờ Sai sau tag mik nx nha

tth Trần Thanh Phương giúp mk vs

\(A=2xy^2\cdot\dfrac{1}{2}x^2y^2x=x^4y^4>0\)(Vì x,y khác 0)