Prove Inequality of $x$ and $y$ with $x^3-y^3=2$ and $x^5-y^5\ge 4$

[anemone]

$x$ and $y$ are two real numbers such that $x^3-y^3=2$ and $x^5-y^5\ge 4$.

Prove that $x^2+y^2\gt 2$.

 

[kaliprasad]

$x$ and $y$ are two real numbers such that $x^3-y^3=2$ and $x^5-y^5\ge 4$.

Prove that $x^2+y^2\gt 2$.

Spoiler

from 1st relation $x > y$ ( if we need to prove refer to 2nd box)
$(x^3-y^3)(x^2+y^2) = x^5 - y^5 + x^3y^2 - x^2y^3 = x^5-y^5 + x^2y^2(x-y) > x^5-y^5$
hence
$(x^2+y^2) > \frac{x^5-y^5}{x^3-y^3} > \frac{4}{2} \,or\,2 $ as $(x^5-y^5),(x^3-y^3)$ both positive

Spoiler

$(a^3-b^3) = (a-b)(a^2+b^2 + ab)$

$a^2+b^2 + ab $ if both same sign then $(a-b)^2+ 3ab$ positive

if opposite signs

then a$a^2+b^2 + ab = (a+b)^2 -ab$ both positive and sum positive

 

[anemone]

Very well done, kaliprasad!(Cool)

 

[Greg]

$x$ and $y$ are two real numbers such that $x^3-y^3=2$ and $x^5-y^5\ge 4$.

Prove that $x^2+y^2\gt 2$.

Spoiler

Note that $x>y$. Proof: $x^3-y^3=(x-y)(x^2+xy+y^2)$. as $x^2+xy+y^2$ is positive and $x^3-y^3$ is positive, it follows that $x>y$.

$$x^5-y^5\ge2(x^3-y^3)$$
$$(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)\ge2(x-y)(x^2+xy+y^2)$$
$$x^4+x^3y+x^2y^2+xy^3+y^4-2(x^2+y^2)-2xy\ge0$$
$$(x^2+y^2)^2-x^2y^2+x^3y+xy^3-2(x^2+y^2)-2xy\ge0$$
$$(x^2+y^2-2)(x^2+y^2)+xy(x^2+y^2-2)-x^2y^2\ge0$$
$$(x^2+y^2-2)(x^2+y^2+xy)-x^2y^2\ge0\quad(1)$$

It follows from $(1)$ that $x^2+y^2>2$.