Question - Proving an Algebraic Inequality

Given \( 0 < p < 1 \)

We need to prove \( 1 - 3p + 4p^3 > 0 \).

Let's rearrange the inequality:

\( 4p^3 - 3p + 1 > 0 \).

Consider the expression as a function \( f(p) = 4p^3 - 3p + 1 \).

Let's find the discriminant \( \Delta \) of the corresponding quadratic equation to \( f'(p) \) to check the number of real roots.

\( f'(p) = 12p^2 - 3 \)

For \( f'(p) = 0 \),

\( 12p^2 - 3 = 0 \)

\( p^2 = \frac{1}{4} \)

\( p = \frac{1}{2} \) or \( p = -\frac{1}{2} \) (reject as \( -\frac{1}{2} < 0 \) which is not in the domain \( 0 < p < 1 \))

Since \( f'(p) \) has only one real root in \( (0,1) \), \( f(p) \) has at most one extremum in \( (0,1) \).

Now, let's check the sign of \( f(p) \) at the found extremum \( p = \frac{1}{2} \):

\( f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right) + 1 = 4\left(\frac{1}{8}\right) - \frac{3}{2} + 1 = \frac{1}{2} - \frac{3}{2} + 1 = 0 \)

As \( 0 < p < 1 \) and since \( f(p) \) is continuous and has no other extremum, \( f(p) \) does not change sign in \( (0,1) \).

Finally, test the endpoints:

\( f(0) = 4(0)^3 - 3(0) + 1 = 1 > 0 \)

\( f(1) = 4(1)^3 - 3(1) + 1 = 2 > 0 \)

Therefore, as \( f(0) > 0 \) and \( f(1) > 0 \) and there are no roots in \( (0,1) \), \( f(p) > 0 \) for all \( p \) in \( (0,1) \), which completes the proof.