Problem source

\begin{questionparts}  
\item
Show that, for $0\le x\le 1$, the largest value of $\displaystyle \frac{x^6}{(x^2+1)^4}$ is $\frac1{16}$.
\item Find constants $A$, $B$, $C$ and $D$ such that, for all $x$, 
\[ \frac{1}{(x^2+1)^4}= \frac{\d \ }{\d x} \left( \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3}\right) +\frac{Dx^6}{(x^2+1)^4}. \] 
\item Hence, or otherwise, prove that 
\[ \frac{11}{24} \le \int_{0}^{1}\frac{1}{(x^{2}+1)^{4}}\, \d x  \le \frac{11}{24} + \frac 1{16} \;  . \] 
\end{questionparts}

Solution source

\begin{questionparts}
\item $x^2 + 1 \geq 2x \Rightarrow \frac{1}{x^2+1} \leq \frac1{2x} \Rightarrow \frac{x^6}{(x^2+1)^4}\leq \frac{x^6}{16x^2} = \frac1{16}x^2 \leq \frac1{16}$ with equality when $x = 1$

\item $\,$ \begin{align*}
&& RHS &=  \frac{\d \ }{\d x} \left( \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3}\right) +\frac{Dx^6}{(x^2+1)^4} \\
&&&= \frac{(5Ax^4+3Bx^2+C)(x^2+1)^3-6x(x^2+1)^2(Ax^5+Bx^3+Cx)}{(x^2+1)^6}+\frac{Dx^6}{(x^2+1)^4} \\
&&&= \frac{(5Ax^4+3Bx^2+C)(x^2+1)-6x(Ax^5+Bx^3+Cx)+Dx^6}{(x^2+1)^4} \\
&&&= \frac{(D-A) x^6 + (5A-3B) x^4 + (3B-5C)x^2 + C}{(x^2+1)^4} \\
\Rightarrow && C &= 1 \\
&& 3B &= 5 \quad\quad\quad\Rightarrow B = \frac53 \\
&& 5A &= 3B = 5\Rightarrow A = 1 \\
&& D &= A \quad\quad \Rightarrow D = 1
\end{align*}
\item So
\begin{align*}
&& I &= \int_{0}^{1}\frac{1}{(x^{2}+1)^{4}}\, \d x  \\
&&&= \int_{0}^{1}\frac{\d \ }{\d x} \left( \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3}\right) +\frac{Dx^6}{(x^2+1)^4} \, \d x \\
&&&= \left [ \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3} \right]_0^1 +  \int_{0}^{1}\frac{Dx^6}{(x^2+1)^4} \, \d x \\
&&& \leq \frac{A+B+C}{8} + \frac1{16} \\
&&&= \frac{2+\frac53}{8} + \frac1{16} = \frac{11}{24} + \frac{1}{16} \\
&& I &\geq  \left [ \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3} \right]_0^1  = \frac{11}{24}
\end{align*}
\end{questionparts}