Với \(x,y>0\). Áp dụng BĐT AM-GM, ta có:

\(x^4+y^2\ge2x^2y\)

\(\Rightarrow x^4+y^2+2xy^2\ge2x^2y+2xy^2=2xy\left(x+y\right)\)

\(\Rightarrow\frac{1}{x^4+y^2+2xy^2}\le\frac{1}{2xy\left(x+y\right)}\)(đpcm)

áp dụng bổ đề     \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)(bạn dùng cô-si,xét tích \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a+b+c\right)\))

\(\Leftrightarrow\frac{1}{x^2+2xy}+\frac{1}{y^2+2yz}+\frac{1}{z^2+2xz}\ge\frac{9}{\left(x+y+z\right)^2}=\frac{9}{1^2}\)

\(=\left(x-y\right)^2+1\ge1>0,\forall x,y\)

\(x^2-2xy+y^2+1\)

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

Vì \(\left(x-y\right)^2\ge0\) với mọi \(x,y\in R\)

\(\Rightarrow\left(x-y\right)^2+1\ge1\) với mọi \(x,y\in R\)

\(\Rightarrow\left(x-y\right)^2+1>0\) với mọi \(x,y\in R\) (đpcm)

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

\(\left(x-y\right)=t\Rightarrow t^2-t+1=t^2-2.\frac{1}{2}t+\frac{1}{4}+\frac{3}{4}=\left(t-\frac{1}{2}\right)^2+\frac{3}{4}>0\)

=>đpcm

\(x^2+y^2-2xy+x-y+1\)

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

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

\(=\left(x-y+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x;y\)

P.s: cách này dễ hiểu hơn cách của Nguyễn Hưng Phát

x² + y² - 2xy + x - y + 1 = (x - y)² + (x - y) + 1

Đặt x - y = t

Ta có: x² + y² - 2xy + x - y + 1 = t² + t + 1 = (t + \(\dfrac{1}{2}\))² + \(\dfrac{3}{4}\) > 0 với mọi t

\(\left(x+y\right)\left(x+y\right)=x^2+xy+xy+y^2=x^2+2xy+y^2\)

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

\(\left(x-z\right)\left(x+z\right)=x^2+xz-xz-z^2=x^2-z^2\)