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

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- Sep 10th 2008, 06:35 PMrmpatel5inequalities
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

- Sep 10th 2008, 06:42 PM11rdc11
For the example assume a = 0 so b>0

$\displaystyle 0^2<0<b^2$

which is false - Sep 10th 2008, 06:58 PMShyam
$\displaystyle a<b$ ........................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)

$\displaystyle a^2<ab<b^2$ - Sep 10th 2008, 07:57 PMacc100jt
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