Show that the function f(x) = (4x (1-x))/((1+x)²) cannot

Show that the function $\displaystyle f(x) = \frac{4x\bigl(1-x\bigr)}{\bigl(1+x\bigr)^{2}}$ cannot take values greater than $\dfrac12$.

Step 1: What has to be proved

The function is defined for $x\neq-1$. Set

\[ \varphi(x) = \frac12-\frac{4x\bigl(1-x\bigr)}{\bigl(1+x\bigr)^{2}}, \] so the claim is that $\varphi(x)\ge0$ throughout the domain.

Step 2: Bring to a common denominator

\[ \varphi(x) = \frac{\tfrac12\bigl(1+x\bigr)^{2}-4x\bigl(1-x\bigr)}{\bigl(1+x\bigr)^{2}} = \frac{\tfrac12+x+\tfrac{x^{2}}{2}-4x+4x^{2}}{\bigl(1+x\bigr)^{2}}. \tag{1} \]

Step 3: The numerator is a perfect square

\[ \frac12+x+\frac{x^{2}}{2}-4x+4x^{2} = \frac12-3x+\frac{9x^{2}}{2} = \frac{1-6x+9x^{2}}{2} = \frac{\bigl(1-3x\bigr)^{2}}{2}. \tag{2} \]

Step 4: When equality holds

$\varphi(x) = 0$ exactly when $1-3x = 0$, that is $x = \dfrac13$, and there

\[ f\!\left(\frac13\right) = \frac{4\cdot\tfrac13\cdot\tfrac23}{\left(\tfrac43\right)^{2}} = \frac{\tfrac89}{\tfrac{16}{9}} = \frac12. \]

So the bound is sharp: $\dfrac12$ is attained, and is the maximum of $f$.

Step 5: Verification

\[f\!\left(\tfrac13\right) = 0.5; \qquad f(0.2) = 0.4444444; \qquad f(0.5) = 0.4444444 \;\;\checkmark\]

\[f(0) = f(1) = 0; \qquad f(2) = -0.8888889; \qquad f(-0.5) = f(-3) = -12 \;\;\checkmark\]

\[\text{By calculus: } f'(x) = \frac{4\bigl(1-3x\bigr)}{\bigl(1+x\bigr)^{3}},\ \text{zero only at } x = \tfrac13 \;\;\checkmark\]