Prove that for any real $(x,y)$, $x^2+|y| \geq 4$ for this equation system

Question

How can I prove that for any real $(x,y)$ root this system of equations

$$x^4 + y^2 = \Big(a + \frac{1}{a}\Big)^3$$

$$x^4 - y^2 = \Big(a - \frac{1}{a}\Big)^3$$

the inequation $x^2+|y| \geq 4$ is always true?

Also, when will $x^2+|y| = 4$ be true for this system?

I think I could make use of the AM–GM inequality, couldn't get anything properly working with it, though.

Answer 1

Hint: $x^4 = \frac {(x^4 + y^2)+(x^4 - y^2)}2$ and $y^2 = \frac {(x^4 + y^2)-(x^4 - y^2)}2$

So $x^4 = a^3 + 3a*\frac 1{a^2}= a^3 + 3\frac 1a$ and $y^2 = 3a^2*\frac 1a + \frac 1a^3 = 3a + \frac 1{a^3}$ (and presumably $a > 0$ else we'd have $\frac 10$ or $x^4 < 0$)

So $x^2 + |y| =\sqrt{a^3 + 3\frac 1a} +\sqrt{3a + \frac 1{a^3}}$ so by AM-GM

$x^2 + |y| \ge 2\sqrt[4]{(a^3 + 3\frac 1a)(3a + \frac 1{a^3})}=2\sqrt[4]{3a^4+9+1+\frac 3{a^4}}$.

And applying the AM-GM result that for $x > 0$ that $x + \frac 1x\ge 2\sqrt{x\frac 1x} = 2$ we have

$x^2 + |y| \ge2\sqrt[4]{3a^4+9+1+\frac 3{a^4}}=2\sqrt[4]{10+3(a^4 +\frac 1{a^4})} \ge 2\sqrt[4]{10 + 3*2}=2\sqrt[4]{16}=2*2=4$.

Answer 2

Hint:

Add and subtract the equations and divide by $2$ to get

$x^4=a^3+\dfrac3a$ and $y^2=\dfrac1{a^3}+3a$.