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

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The discussion centers on proving the inequality $x^2 + y^2 \ge 2$ under the constraints $x^3 - y^3 = 2$ and $x^5 - y^5 \ge 4$. Participants explore various mathematical approaches to establish this relationship, emphasizing the implications of the given equations. The conversation highlights the importance of manipulating algebraic expressions and applying inequalities effectively. Ultimately, the goal is to demonstrate that the conditions lead to the desired inequality. The proof remains a focal point of the discussion, showcasing the interplay between the equations and the inequality.
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Given $x^3-y^3=2$ and $x^5-y^5\ge 4$ for all real $x$ and $y$. Prove that $x^2+y^2\ge 2$.
 
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anemone said:
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:

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: $x^2+y^2 \geq 2.$
 
Last edited:
Thanks lfdahl for participating and thanks too for the solution! Good job! :D

My solution:

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 $x^2+y^2\ge 2$.
 
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