60. Proof: Without loss of generality, let $a \geqslant b$, then $a^{2}+a b \geqslant b^{2}+a b$. Now we prove $a^{2}+a b>2$. If $a^{2}+a b \leqslant 2$, then

and
$$b \leqslant \frac{2-a^{2}}{a}$$

Therefore
$$31, b>1$$

Thus
$$$$11$, this contradicts the assumption$a>b$. Hence$a^{2}+a b>2$.