Inequality for quartic polynomial depending on a parameter Let $f(x) = \frac 14 x^4 - \frac \alpha2 x^2 - (\alpha-1) x - \frac \alpha 2 + \frac 3 4 $. 
I want to show that there exists an $\alpha>1$ such that $f(x)\geq 0 $ for $x\leq 0$. Even more, it would be nice to exhibit explicitly such an $\alpha$.  
I would like to avoid to look at the zeros of f(x) or f'(x), which seems not so handy because the formulas for solution of quartic/cubic equations are complicated.
Maybe some continuity argument? (Note that for $\alpha=1$ the inequality is true, the minimum on $x\leq 0$ is reached at $x=-1$ and equals $0$ ) .
Thanks
 A: You need
$$x \le 0 \implies f(x) = \frac14 (x+1)^2 \left((x-1)^2+2-2\alpha \right) \ge 0 $$
So it is sufficient to ensure $(x-1)^2 + 2 \ge 2\alpha$, which will be true for $\alpha \le \frac32$.  For e.g. with $\alpha = \frac32$, you have $f(x) = \frac14 x (x-2)(x+1)^2$ which is non-negative for all negative $x$.
A: Note that $f(0)=-\frac{\alpha}{2}+\frac{3}{4}$ then $1<\alpha\leq\frac{3}{2}$
also:
$$f'(x)=(x+1)[x^2-x+1-\alpha]$$ and $$f''(x)=3x^2-\alpha$$
you can choice $\alpha=\frac{3}{2}$, because the fuction reach minimun in $x=-1$, maximun in $x=\frac{1-\sqrt{3}}{2}$, is decreasing in $(-\infty,-1)$ increasing in $(-1,\frac{1-\sqrt{3}}{2})$ and decreasing in $(\frac{1-\sqrt{3}}{2},0)$ and $f(-1)=0$, $f(0)=0$.
A: All values of $1.5\geq\alpha > 1$ satisfy your requirements.
Looking at other answers here, this might be a case of nuking a fly. But I'll post it anyway.
Edit: Heres a plot of the function with $\alpha$ going from 1 to 1.5

The polynomial has a double root at $x=-1$, since:
$$f(x)=\frac 14 x^4 - \frac \alpha2 x^2 - (\alpha-1) x - \frac \alpha 2 + \frac 3 4$$
simplifies to $0$ at $x=-1$ and
$$f'(x)=x^3-\alpha x-\alpha+1$$
also simplifies to $0$ at $x=-1$.
This means that $-1$ is either a local minimum or a local maximum of the function.
To find out if it's a minimum or maximum, we must first prove that there are no  other minimums/maximums that are smaller than $-1$
This means finding the remaining roots of the functions derivative:
$$f'(x)=x^3-\alpha x-\alpha+1$$
Which, except for $-1$ also has roots
$$\frac{1}{2}(1-\sqrt{4\alpha-3})$$
$$\frac{1}{2}(1+\sqrt{4\alpha-3})$$
Since $\alpha>1$ then $\sqrt{4\alpha-3}>1$. This means that the smaller of those roots is $\frac{1}{2}(1-\sqrt{4\alpha-3})$ and since $\frac{1}{2}(1-\sqrt{4\alpha-3})>-1$ for $3\geq\alpha$. Conclusion here is that $-1$ is the smallest minimum/maximum for $1.5\geq\alpha>1$.
This means that if $f(x)>0$ for any $x<-1$.
We now take $x=-2$, and see that $f(-2)=\frac{1}{4}(11-2\alpha)$
This is a decreasing function, but it is positive in the desired interval $\alpha\in(1,1.5]$
We now know that $f(x)\geq0$ for $x\leq-1$, all that is remaining is to prove for the interval $(-1,0)$.
We know that $-1$ is a minimum, therefor the values from $-1$ to the next larger root must be positive.
Out goal is to show that $f(x)$ with $1.5\geq\alpha>1$ have no roots in the open interval $(-1,0)$.
The roots of $f(x)$ in terms of $\alpha$ is the following (excluding the already known $-1$, and not important in this case)
$$1-\sqrt{2}\sqrt{\alpha-1}$$
$$1+\sqrt{2}\sqrt{\alpha-1}$$
The root $1+\sqrt{2}\sqrt{\alpha-1}$ is clearly positive for $\alpha>1$, so that root will never be in the interval $(-1,0)$.
The root $1-\sqrt{2}\sqrt{\alpha-1}$ becomes negative at some point, if we can show that $1-\sqrt{2}\sqrt{\alpha-1}\geq0$ for $1.5\geq\alpha > 1$. Then we are done.
$\sqrt{2}\sqrt{\alpha-1}$ is an increasing function for $\alpha\geq1$, so $1-\sqrt{2}\sqrt{\alpha-1}$ must be a decreasing function. Let's solve for x.
$$1-\sqrt{2}\sqrt{\alpha-1}=0$$
$$\sqrt{2}\sqrt{\alpha-1}=1$$
$$\sqrt{\alpha-1}=\frac{1}{\sqrt{2}}$$
$$\alpha-1=\frac{1}{2}$$
$$\alpha=\frac{3}{2}$$
The root of the function is $1.5$, after that value the root is negative. But the root must only be nonnegative for $1.5\geq\alpha>1$.
We are done.
