\(a^2+b^2+3>ab+a+b\)

\(\Leftrightarrow2\left(a^2+b^2+3\right)>2\left(ab+a+b\right)\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(a^2-2ab+b^2\right)+4>0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(a-b\right)^2+4>0\) \(\forall a,b\)

Vậy \(a^2+b^2+3>ab+a+b\forall a,b\)

\(n_{O_2}=\dfrac{m}{M}=\dfrac{6,4}{32}=0,2\left(mol\right)\)

\(4Al\)   \(+\)   \(3O_2\)   →   \(2Al_2O_3\)

               \(0,2\)    →    \(\dfrac{2}{15}\)              \(\left(mol\right)\)

\(m_{Al_2O_3}=n.M=\dfrac{2}{15}.102=13,6\left(g\right)\)

\(\dfrac{-5}{7}-\dfrac{1}{3}=\dfrac{-15}{21}-\dfrac{7}{21}=\dfrac{-22}{21}\)