<s>[INST] 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...
 [/INST] $x^2+y^2>1$, and $x^4+y^4 <1$.
Consider $y=x$ in the first quadrant.
$2x^2 >1$, and $2x^4 <1;$
$x ^2 >1/2$ , and $x^2 <1/√2$, i .e.
$1/2=0.5  < x^2 < √2/2=0.7071..$
Points on $y=x$ , line through origin, with
$1<x^2+y^2 < √2 $, satisfy
$x^4+y^4 <1$.
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