What is the number of pairs of $(a,b)$ of positive real numbers satisfying $a^4+b^4<1$ and $a^2+b^2>1$? 
The number of pairs of $(a,b)$ of positive real numbers satisfying
  $a^4+b^4<1$ and $a^2+b^2>1$ is -
$(i)$0
$(ii)$1
$(iii)$2
$(iv)$ more than 2

Solution:We have  $a,b>0$,
According to the given situation,$0<a^4+b^4<1<a^2+b^2\implies a^4+b^4<a^2+b^2\implies 0<a^2(a^2-1)<b^2(1-b^2)\implies b^2(1-b)(1+b)<0 ;a^2(a-1)(a+1)>0\implies a\in(-\infty,-1)\cup (0,1);b\in(-1,0)\cup(1,\infty)$
But, in particular, if we choose $a=-1/2,b=1/2$,then $a^4+b^4=1/16+1/16=1/8<1$ and $a^2+b^2=1/4+1/4=1/2<1$.Which contradicts the given condition $a^2+b^2>1$
Where is the mistake in my approach? How should I approach this problem(means how to think ?). Can Triangle inequality be used here?
Please provide some hint...

Let $a^3+b^3=1$, there are uncountably many such pairs with $a\in (0,1)$ as we can set $a^3=\cos^2t, b^3=\sin^2t$. 
Now for all these pairs, $a^4+b^4<a^3+b^3=1<a^2+b^2$ is obvious. 
