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anemone
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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 $x^2+y^2\ge 2$.
anemone said:Given $x^3-y^3=2$ and $x^5-y^5\ge 4$ for all real $x$ and $y$. Prove that $x^2+y^2\ge 2$.
The given equations, $x^3-y^3=2$ and $x^5-y^5\ge 4$, can be used to show that $x^2+y^2\ge 2$ by manipulating them algebraically. We can start by rearranging the first equation to get $x^3-y^3=2$ as $(x-y)(x^2+xy+y^2)=2$. Then, we can substitute $(x^2+xy+y^2)$ with $\frac{2}{x-y}$ from the second equation, giving us $\frac{x^2 + xy + y^2}{x-y} \ge 2$. Simplifying this further, we get $x^2 + y^2\ge 2$, which proves the given statement.
The given equations, $x^3-y^3=2$ and $x^5-y^5\ge 4$, provide a relationship between the variables $x$ and $y$. By manipulating these equations, we can arrive at a statement that involves $x^2$ and $y^2$, which is what we are trying to prove. This shows that the given equations are useful in proving the given statement.
Yes, there are other ways to prove $x^2+y^2\ge 2$ without using the given equations. One way is to use the Cauchy-Schwarz inequality, which states that for any two real numbers $a$ and $b$, $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$. We can apply this inequality to the given statement by letting $a=x$, $b=y$, $c=1$, and $d=1$, which gives us $(x^2+y^2)(1^2+1^2)\ge (x+y)^2$. Simplifying this, we get $x^2+y^2\ge 2xy$. Then, by substituting $xy$ with $\frac{x^2+y^2}{2}$ from the given statement, we get $x^2+y^2\ge 2$, proving the statement.
Yes, the given equations can be used to prove other inequalities. By manipulating the equations algebraically, we can arrive at statements that involve different powers of $x$ and $y$, such as $x^3+y^3\ge 2$ or $x^4+y^4\ge 2$, depending on the specific inequality that we want to prove.
Yes, there are restrictions on the values of $x$ and $y$ for the given equations to be valid. From the first equation, $x^3-y^3=2$, we can see that $x\neq y$ since the denominator cannot be equal to zero. Additionally, from the second equation, $x^5-y^5\ge 4$, we can see that $x\ge y$ since the numerator must be greater than or equal to the denominator for the inequality to hold. Therefore, the values of $x$ and $y$ must satisfy both of these conditions for the given equations to be valid.