The minimum value of $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x, 0\le x\le \frac{\pi }{2}$ is

We have,

$\begin{array}{l} f\left(x\right)={\sin }^{4}x+{\cos }^{4}x,0\le x\le \frac{\pi }{2}\\ \Rightarrow f\left(x\right)={\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-\frac{1}{2}{\sin }^{2}2x\\ \Rightarrow f\left(x\right)=1-\frac{1}{4}(1-\cos 4x)\\ \Rightarrow f\left(x\right)=\frac{3}{4}+\frac{1}{4}\cos 4x\end{array}$

$\because -1\le \cos 4x\le 1\text{ for }x\in [0,\pi /2]$

$\therefore -\frac{1}{4}\le \frac{1}{4}\cos 4x\le \frac{1}{4}$ for all $x\in \left[0, \pi /2\right]$

$\Rightarrow \frac{1}{2}\le \frac{3}{4}+\frac{1}{4}\cos 4x\le 1$for all $x\in \left[0, \pi /2\right]$

$\Rightarrow \frac{1}{2}\le f\left(x\right)\le 1$ for all $x\in \left[0, \pi /2\right]$