Proving an Algebraic Inequality
<p>Given \( 0 < p < 1 \)</p>
<p>We need to prove \( 1 - 3p + 4p^3 > 0 \).</p>
<p>Let's rearrange the inequality:</p>
<p>\( 4p^3 - 3p + 1 > 0 \).</p>
<p>Consider the expression as a function \( f(p) = 4p^3 - 3p + 1 \).</p>
<p>Let's find the discriminant \( \Delta \) of the corresponding quadratic equation to \( f'(p) \) to check the number of real roots.</p>
<p>\( f'(p) = 12p^2 - 3 \)</p>
<p>For \( f'(p) = 0 \),</p>
<p>\( 12p^2 - 3 = 0 \)</p>
<p>\( p^2 = \frac{1}{4} \)</p>
<p>\( p = \frac{1}{2} \) or \( p = -\frac{1}{2} \) (reject as \( -\frac{1}{2} < 0 \) which is not in the domain \( 0 < p < 1 \))</p>
<p>Since \( f'(p) \) has only one real root in \( (0,1) \), \( f(p) \) has at most one extremum in \( (0,1) \).</p>
<p>Now, let's check the sign of \( f(p) \) at the found extremum \( p = \frac{1}{2} \):</p>
<p>\( f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right) + 1 = 4\left(\frac{1}{8}\right) - \frac{3}{2} + 1 = \frac{1}{2} - \frac{3}{2} + 1 = 0 \)</p>
<p>As \( 0 < p < 1 \) and since \( f(p) \) is continuous and has no other extremum, \( f(p) \) does not change sign in \( (0,1) \).</p>
<p>Finally, test the endpoints:</p>
<p>\( f(0) = 4(0)^3 - 3(0) + 1 = 1 > 0 \)</p>
<p>\( f(1) = 4(1)^3 - 3(1) + 1 = 2 > 0 \)</p>
<p>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.</p>
