Non negativity condition for polynomials? Let $p(x)=3ax^2+2bx+(1-a-b), \,\,\,\,0<x<1$
What are the conditions on a and b to $p(x)>0$
Could someone give me a hint ?
 A: Hint: complete the square. This means writing $p(x)$ in the form
$$p(x)=A(x+B)^2+C
$$
where $A$, $B$, $C$ are constants possibly depending on $a$ and $b$, but not on $x$. When you're done you will have expressed $p(x)$ as a nonnegative piece $A(x+B)^2$ (this requires $A\ge0$) and a constant $C$. So if we require $p(x)>0$ for all $x$ we must have $A\ge0$ and $C>0$.
In your case you'd complete the square by starting out
$$
P(x)=3a(x^2+\frac{2b}{3a}x)+(1-a-b) .
$$
To get this first section in the form $A(x+B)^2$ you'd want to complete the square with $(\frac b{3a})^2$. To repair what you've just done, you need to subtract it off again:
$$P(x)=3a\left(x^2+\frac{2b}{3a}x + (\frac b{3a})^2 - (\frac b{3a})^2)\right) + (1-a-b).$$
Next, regroup:
$$\begin{align}
P(x)&=3a\left(x^2+\frac{2b}{3a}x + (\frac b{3a})^2\right) -3a(\frac b{3a})^2 + (1-a-b).\\
&=3a\left(x+\frac b{3a}\right)^2 -3a(\frac b{3a})^2 + (1-a-b).\\
\end{align}
$$
From this last expression you can read off $A$, $B$, $C$. (You can simplify $C$ down to a more tidy expression.)
A: Hint: 


*

*You must have $3a>0$

*The derivative is $6ax+2b$ vanishes at $x = -{{b}\over{3a}}$ which is the minimum if $a>0$.

*Find $a,b$ such that $p(-{b\over{3a}})>0$.
