Dễ thấy:

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

Áp dụng Cô-si:

     \(\dfrac{3\left(x+y\right)^2}{4}+1\ge2\sqrt{\dfrac{3\left(x+y\right)^2}{4}.1}=\sqrt{3}\left|x+y\right|\ge\sqrt{3}\left(x+y\right)\)

Do đó:

     \(\left(x+y\right)^2+1-xy\ge\sqrt{3}\left(x+y\right),\forall x,y\in R\)

a/ \(x^2+xy+y^2+1=\left(x^2+xy+\frac{y^2}{4}\right)+\frac{3y^2}{4}+1=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1>0\)

b/ \(x^2+5y^2+2x-4xy-10y+14\)

\(=\left(x^2-4xy+4y^2\right)+2\left(x-2y\right)+1+\left(y^2-6y+9\right)+4\)

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

\(=\left(x-2y+1\right)^2+\left(y-3\right)^2+4>0\)

áp dụng BĐT cô si cho 2 số ta có

\(\dfrac{x}{y}+\dfrac{y}{x}\ge2\sqrt{\dfrac{x}{y}.\dfrac{y}{x}}\)

⇔ \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\left(đpcm\right)\)

Cách khác:

Đặt \(A=\dfrac{x}{y}+\dfrac{y}{x}\)

\(A=\dfrac{x^2+y^2}{xy}\)

Lại có:\(\left(x-y\right)^2\ge0\)

\(\Rightarrow x^2-2xy+y^2\ge0\)

\(\Leftrightarrow x^2+y^2\ge2xy\)

\(\Rightarrow A=\dfrac{x^2+y^2}{xy}\ge\dfrac{2xy}{xy}=2\left(đpcm\right)\)

Dấu "=" xảy ra khi x=y