# If $\gamma = ad(b\beta-e\alpha)$, $\delta=4abcde^2\beta^2$, and $e<\frac{b\beta}{\alpha}$, is $\gamma^2=\delta$?

Suppose we have the quantities $a, b, c, d, \alpha, \beta>0$. Also note that $$e<\frac{b\beta}{\alpha}.$$

Let $\gamma = ad(b\beta-e\alpha)$ and $\delta=4abcde^2\beta^2$. Can anyone show $\gamma^2=\delta$ is not always necessarily true using the above information?

It's not true.

Take $b=2$, and $a=d=e=\alpha=\beta=1$, and $c=\frac{1}{8}$.

Then $\gamma^2=\delta=1$.

Edit:

For an example of when $\gamma^2=\delta$ is not necessarily true, take $b=2$, and $a=d=e=\alpha=\beta=1$, and $c$ some number different from $\frac{1}{8}$.

• So $\gamma^2=\delta$ for every $a,b,c,d,e,\alpha, \beta$? I have edited my question slightly. – user2850514 Nov 8 '15 at 19:21
• @user2850514 no, far from. For most choices of your variables, you have that $\gamma^2 \neq \delta$, but I showed you that it does not hold for all choices. – Mankind Nov 8 '15 at 19:23
• I have modified the question slightly. – user2850514 Nov 8 '15 at 19:24
• @user2850514 I have edited my answer. – Mankind Nov 8 '15 at 19:26