Minimum value

Take a= (sin x)^2 & b=(cos x)^2. (since a+b=1)
a^4+b^4=(sin x)^8+( cos x)^8.
Differentiating it we get the minima at (sin x)^2=(cos x)^2= $\frac{1}{2}$ (Since sin x,cos x>0)
ANSWER= $\frac{1}{16}$ + $\frac{1}{16}$ = $\frac{1}{8}$ =0.125

This is a classical inequality problem: By AM-GM inequality

$\frac{a^2+b^2}{2}$ $\geq$ $ab$

$\Rightarrow$ $a^2+b^2$ $\geq$ $2ab$

$\Rightarrow$ $a^2+2ab+b^2$ $\geq$ $4ab$

$\Rightarrow$ $(a+b)^2$ $\geq$ $4ab$

$\Rightarrow$ $ab$ $\leq$ $\frac{1}{4}$

$\Rightarrow$ $2(ab)^2$ $\leq$ $2 \times \frac{1}{16}\$ = $\frac{1}{8}$ .....(i)

Now, consider the following inequality (ByAM-GM inequality)

$a^2+b^2$ $\geq$ $2ab$

$\Rightarrow (a^2+b^2)^2 \geq 4(ab)^2$

$\Rightarrow a^4+b^4+2(ab)^2 \geq 4(ab)^2$

$\Rightarrow a^4+b^4 \geq 2(ab)^2$

$\Rightarrow a^4+b^4 \geq \frac{1}{8}$ From(i),

Thus, the minimum value occurs when the expresion is equal to $\frac{1}{8}$ = $0.125$