0≤a<b, prove that a^2≤ab<b^2. Also, show by example that it does not follow that a^2<ab<b^2
$\displaystyle a<b$ ........................eqn(1)
Multiply both sides with a, we get,
$\displaystyle a^2<ab$ .....................eqn(2)
Now multiply both sides of eqn(1) with b, we get
$\displaystyle ab<b^2$ ...................eqn (3)
from eqn(2) and eqn (3)
$\displaystyle a^2<ab<b^2$
Multiply both sides by $\displaystyle a$
$\displaystyle a<b \Rightarrow a^{2}\le ab$ since $\displaystyle a$ can be 0
Multiply both sides by $\displaystyle b$
$\displaystyle a<b \Rightarrow ab<b^{2}$
Hence $\displaystyle a^{2}\le ab<b^{2}$
A counter example for $\displaystyle a^{2}<ab<b^{2}$ is given by 11rdc11