Proving $x^2+y^2\ge 2$ with Given $x^3-y^3=2$ and $x^5-y^5\ge 4$

[anemone]

Given $x^3-y^3=2$ and $x^5-y^5\ge 4$ for all real $x$ and $y$. Prove that $x^2+y^2\ge 2$.

 

[lfdahl]

Given $x^3-y^3=2$ and $x^5-y^5\ge 4$ for all real $x$ and $y$. Prove that $x^2+y^2\ge 2$.

My solution:

Spoiler

Note, that

\[x^5-y^5 = (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)\geq 4 \\\\ x^3-y^3 =(x-y)(x^2+xy+y^2)=2\]

Under the condition $x \ne y$:

\[\frac{x^5-y^5}{x^3-y^3}=\frac{x^4+x^3y+x^2y^2+xy^3+y^4}{x^2+xy+y^2}= \frac{(x^2+xy+y^2)^2-x^3y-2x^2y^2-xy^3}{x^2+xy+y^2} \\\\ \\\\ =x^2+xy+y^2-\frac{xy(x^2+2xy+y^2)}{x^2+xy+y^2} =x^2+y^2-\frac{x^2y^2}{x^2+xy+y^2}\geq 2\]

All I need to show is that:

\[\frac{x^2y^2}{x^2+xy+y^2}\geq 0\]

This holds iff $x^2+xy+y^2 > 0$:

\[x^2+xy+y^2 = (x+\frac{y}{2})^2-\frac{1}{4}y^2 +y^2 =(x+\frac{y}{2})^2+(\frac{\sqrt{3}}{2}y)^2 > 0.\]

Thus: $x^2+y^2 \geq 2.$

 

[anemone]

Thanks lfdahl for participating and thanks too for the solution! Good job! :D

My solution:

Spoiler

Note that $x^3-y^3=2$ tells us $x\gt y$. Now, considering the product of $(x^3-y^3)(x^2+y^2)$, we have:

$\begin{align*}(x^3-y^3)(x^2+y^2)&=x^5+x^3y^2-x^2y^3-y^5\\&=(x^5-y^5)-x^2y^2(y-x)\end{align*}$

$2(x^2+y^2)+x^2y^2(y-x)=x^5-y^5$

$2(x^2+y^2)+x^2y^2(y-x)\ge 4$

Since $x\gt y$, the term $x^2y^2(y-x)\lt 0$ and so $x^2+y^2\ge 2$.