How to show the following $\alpha x − \beta x^2 - \dfrac{\alpha^2}{4\beta} \leq 0$

Question

I would like to show that the following is true, for all $\alpha \geq 0 , \beta> 0$, we have that:

$$\alpha x − \beta x^2 - \dfrac{\alpha^2}{4\beta} \leq 0, \forall x \in \mathbb{R}$$

Is there a way to prove this result?

Attempt:

$$ 0\leq \beta x^2 - \alpha x + \dfrac{\alpha^2}{4\beta} , \forall x \in \mathbb{R}$$

We want to show that the polynomial on the right hand side only has imaginary roots

Dividing $\beta$ across, yields:

$$ 0\leq x^2 - \dfrac{\alpha}{\beta} x + \dfrac{\alpha^2}{4\beta^2} , \forall x \in \mathbb{R}$$

Then using quadratic formula:

$$\dfrac{1}{2}\left[\dfrac{\alpha}{\beta} \pm \sqrt{\dfrac{4\alpha^2}{4\beta^2} - (4)\dfrac{\alpha^2}{4\beta^2}}\right]$$

which is equal to $$\dfrac{1}{2}\left[\dfrac{\alpha}{\beta}\right]$$

It seems that the roots are real. Therefore this inequality cannot hold true. Can anyone provide assistance?

Answer 1

\begin{align*} -\beta x^{2}+\alpha x-\dfrac{\alpha^{2}}{4\beta}&=-\beta\left(x-\dfrac{\alpha}{2\beta}\right)^{2}+\beta\cdot\dfrac{\alpha^{2}}{4\beta^{2}}-\dfrac{\alpha^{2}}{4\beta}\\ &=-\beta\left(x-\dfrac{\alpha}{2\beta}\right)^{2}\\ &\leq 0. \end{align*}

Answer 2

Just:  $\alpha x − \beta x^2 - \dfrac{\alpha^2}{4\beta} =\dfrac{1}{4 \beta}\big(4\alpha\beta x − 4\beta^2 x^2 - \alpha^2\big) = -\, \dfrac{1}{4 \beta}\big(2\beta x - \alpha\big)^2\leq 0\,$.

Answer 3

Using the famous inequality: $2AB \le A^2+B^2$ with $A = x\sqrt{\beta}, B = \dfrac{\alpha}{2\sqrt{\beta}}$ the conclusion follows.

Answer 4

By your work we need to prove that: $$ x^2 - \dfrac{\alpha}{\beta} x + \dfrac{\alpha^2}{4\beta^2}\geq0,$$ which is $$\left(x-\frac{\alpha}{2\beta}\right)^2\geq0,$$ which is obvious.