Let R be the region bounded by $y{x}^4-1+y\le 0,y\ge 0$ and  $0\le x\le 1$. If  $S_R$ be the area of region R, then which of the following statement(s) is (are) TRUE?

$y\le \frac{1}{x^4+1}, x\le 1,y\ge 0$, so region R is shaded in
given figure.
$\frac{dy}{dx}=\frac{-4x^3}{{(x^4+1)}^2}\downarrow$

$\frac{d^{2}y}{d{x}^2}=0$ at  $x={\left(\frac{3}{5}\right)}^{\frac{1}{4}}$ (inflection)
$\left(A\right)S_R\ge$ sum of area of rectangles OYPZ and TGXZ
$\ge \frac{5}{8}\sqrt{\sqrt{\frac{3}{5}}}+\frac{1}{2}(1-\sqrt{\sqrt{\frac{3}{5}}})$

$\ge \frac{1}{8}\sqrt{\sqrt{\frac{3}{5}}}+\frac{1}{2}$

(B)  $S_R\le$ Sum of area of rectangles OQNZ and PMXZ
 $$\begin{gathered} \le \sqrt{\sqrt{\frac{3}{5}}}+\frac{5}{8}(1-\sqrt{\sqrt{\frac{3}{5}}}) \\ \le \frac{5}{8}+\frac{3}{8}\sqrt{\sqrt{\frac{3}{5}}} \end{gathered}$$
(C)  $S_R\ge$area of trapezium OQPZ
$$\begin{gathered} \ge \frac{1}{2}(1+\frac{5}{8}).\sqrt{\frac{3}{5}} \\ \ge \frac{13}{16}\sqrt{\sqrt{\frac{3}{5}}} \end{gathered}$$

$$\begin{gathered} \left(D\right) \because S_R=\underset{0}{\overset{1}{\int }}\frac{dx}{x^4+1} \\ \ge \underset{0}{\overset{1}{\int }}\frac{dx}{x^2+1}\ge \frac{\pi }{4} \end{gathered}$$