What is the number of pairs of $(a,b)$ of positive real numbers satisfying $a^4+b^4<1$ and $a^2+b^2>1$?

Question

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...

Answer 1

Since the inequalities are strict, options (ii) and (iii) are ruled out by general notions of continuity: As soon as you have a single solution $(a,b)$ in the first quadrant, you have an entire open ball of solutions of some (possibly very small) radius. So it suffices to exhibit a single solution. The values $a=b=0.8$ do the trick: $0.8^2=0.64\gt0.5$, so $a^2+b^2\gt1$ but $0.8^4=0.64^2\lt0.7^2=0.49\lt0.5$, so $a^4+b^4\lt1$.

If the argument invoking continuity seems too advanced, a little extra work shows that $(a,b)=(0.8,0.8)$, $(0.8,0.7)$, and $(0.7,0.8)$ already give more than two solutions.

As for the OP's question, "Where is the mistake in my approach?," there are multiples errors in it. Each of the following implications is incorrect:

$$\begin{align} a^4+b^4\lt a^2+b^2&\implies 0\lt a^2(a^2-1)\\ 0\lt b^2(1-b^2)&\implies b^2(1-b)(1+b)\lt0\\ b^2(1-b)(1+b)\lt0&\implies b\in(-1,0)\cup(1\infty)\\ a^2(a-1)(a+1)\lt0&\implies a\in(-\infty,-1)\cup(0,1) \end{align}$$

But even if these implications were correct (or were replaced with something that is correct), the "contradiction" the OP obtains with a particular choice of values at the end of the string of implications is not a contradiction, because the implications only go in one direction (as pointed out in a comment by Hagen von Eitzen). I.e., $"\implies"$ is not the same as $"\iff"$. (Note also that the problem calls for pairs of positive real numbers, so it's inappropriate to $a=-1/2$ anyway.)

Answer 2

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.

Answer 3

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.

Answer 4

Geometrically, the solution set is the set of points outside of the circle $x^2+y^2=1$ but inside the superellipse $x^4+y^4=1$. There are an infinite number of such points.

Answer 5

$x^2+y^2>1$, 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

$x^4+y^4 <1$.