<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] There are infinitely many solutions:
In fact for every $-1<a<1$ , $a\not = 0$ there exist infinitely many $b$'s such that the pair $(a,b)$ satisfies the condition.
Indeed, the conditions can be written as $b^4<1-a^4$ and $b^2>1-a^2$.
Therefore if $b$ satisfies $b<\sqrt[4]{1-a^4}, b>\sqrt[2]{1-a^2}$ then the pair $(a,b)$ satisfy the conditions in the question. (note that here we use the fact that $-1<a<1$ which implies that $1-a^4$ and $1-a^2$ are positive).
It is left to show that $\sqrt[2]{1-a^2}< \sqrt[4]{1-a^4}$. As this would imply that there are infinitely many $b$'s in between those two values.
Since $-1<a<1$ the terms $1-a^2,1-a^4$ are positive so I can square twice the above inequality.
$$(1-a^2)^2 <(1-a^4) $$
Since $(1-a^2)^2 = 1-2a^2 + a^4$ it is enough to show that
$$ 1-2a^2 + a^4 < 1-a^4$$
but this is equivalent to $$a^4<a^2$$
Finally, since $-1<a<1$ and $a\not = 0$ the assertion that $a^2>a^4$ holds.
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