a/ x² + xy + y² + 1

= [x² + 2.x.\(\dfrac{y}{2}\) + (\(\dfrac{y}{2}\) )² ] + \(\dfrac{3y^2}{4}\) + 1

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

Vì \(\left(x+\dfrac{y}{2}\right)^2\) \(\ge\) 0 với mọi x;y

và \(\dfrac{3y^2}{4}\ge0\) với mọi x;y

=> \(\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}\ge0\) với mọi x;y

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

Bien doi tuong BDT 

\(\frac{x^2}{x-2}\ge8\)

\(\Leftrightarrow x^2\ge8x-16\)

\(\Leftrightarrow x^2-8x+16\ge0\)

\(\Leftrightarrow\left(x-4\right)^2\ge0\left(True\right)\)

Dau '=' xay ra khi \(x=4\)

BĐT Cosi cho 2 số a,b >0: 
a + b >= 2căn(ab) 

di từ: ( √a - √b)² ≥ 0 ( voi moi a , b ≥ 0 ) 

<=> a + b - 2√(ab) ≥ 0 

<=> a + b ≥ 2√(ab) 
dau "=" xay ra khi √a - √b = 0 <=> a = b 

Ta có:\(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)\(\forall a,b\ge0\)

\(\Leftrightarrow a+b-2\sqrt{ab}\ge0\)

\(\Leftrightarrow a+b\ge2\sqrt{ab}\left(đpcm\right)\)

\(a+b\ge2\sqrt{ab}\)

\(\Leftrightarrow a+b-2\sqrt{ab}\ge0\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge o\)(Luôn đúng)