Brock Byrd

2022-06-29

How can I prove that for any real $\left(x,y\right)$ root this system of equations
${x}^{4}+{y}^{2}=\left(a+\frac{1}{a}{\right)}^{3}$
${x}^{4}-{y}^{2}=\left(a-\frac{1}{a}{\right)}^{3}$
the inequation ${x}^{2}+|y|\ge 4$ is always true?
Also, when will ${x}^{2}+|y|=4$ be true for this system?

Keegan Barry

Expert

Hint: ${x}^{4}=\frac{\left({x}^{4}+{y}^{2}\right)+\left({x}^{4}-{y}^{2}\right)}{2}$ and ${y}^{2}=\frac{\left({x}^{4}+{y}^{2}\right)-\left({x}^{4}-{y}^{2}\right)}{2}$
So ${x}^{4}={a}^{3}+3a\ast \frac{1}{{a}^{2}}={a}^{3}+3\frac{1}{a}$ and ${y}^{2}=3{a}^{2}\ast \frac{1}{a}+{\frac{1}{a}}^{3}=3a+\frac{1}{{a}^{3}}$ (and presumably $a>0$ else we'd have $\frac{1}{0}$ or ${x}^{4}<0$)
So ${x}^{2}+|y|=\sqrt{{a}^{3}+3\frac{1}{a}}+\sqrt{3a+\frac{1}{{a}^{3}}}$ so by AM-GM
${x}^{2}+|y|\ge 2\sqrt[4]{\left({a}^{3}+3\frac{1}{a}\right)\left(3a+\frac{1}{{a}^{3}}\right)}=2\sqrt[4]{3{a}^{4}+9+1+\frac{3}{{a}^{4}}}$
And applying the AM-GM result that for $x>0$ that
$x+\frac{1}{x}\ge 2\sqrt{x\frac{1}{x}}=2$ we have
${x}^{2}+|y|\ge 2\sqrt[4]{3{a}^{4}+9+1+\frac{3}{{a}^{4}}}=2\sqrt[4]{10+3\left({a}^{4}+\frac{1}{{a}^{4}}\right)}\ge 2\sqrt[4]{10+3\ast 2}=2\sqrt[4]{16}=2\ast 2=4$

racodelitusmn

Expert

Hint:
Add and subtract the equations and divide by $2$ to get
${x}^{4}={a}^{3}+\frac{3}{a}$ and ${y}^{2}=\frac{1}{{a}^{3}}+3a$

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