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Ask for hint for problem of inequality

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,689
Problem:

Let x and y be positive real numbers satisfying the inequality
$\displaystyle x^3+y^3\le x-y$.

Prove that $\displaystyle x^2+y^2\le 1$ .

Hi all, I'm at my wit's end to prove the question as stated above, and I know it's obvious that $\displaystyle x-y>0$ and $\displaystyle x\ne1,\,y\ne1$, and if I factorized the LHS of the given inequality, I get:

$\displaystyle (x+y)(x^2-xy+y^2)\le x-y$

Now, since $\displaystyle x+y>0$, divide the left and right side by $\displaystyle x+y$ to get:

$\displaystyle x^2-xy+y^2\le\frac{x-y}{x+y}$

$\displaystyle x^2+y^2\le\frac{x-y}{x+y}+xy$

$\displaystyle x^2+y^2\le\frac{x-y+xy(x+y)}{x+y}$

And up to this point, I see no credible path to finish what I've started and I must be missing something very important here...
:mad:

As usual, any guidance or help with this problem would be much appreciated.
(Smile)

Thanks.

 

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
\frac{x - y}{x + y} \leq 1
$$
So $x^2 + y^2 \leq 1 + xy$ .

Theorem:
Given real numbers $a = x^2 + y^2$ and $b = 1$ such that
$$
a\leq b + \varepsilon,\quad \forall\varepsilon > 0.
$$
Then $a\leq b$.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,689
$$
\frac{x - y}{x + y} \leq 1
$$
So $x^2 + y^2 \leq 1 + xy$ .

Theorem:
Given real numbers $a = x^2 + y^2$ and $b = 1$ such that
$$
a\leq b + \varepsilon,\quad \forall\varepsilon >0
$$
Then $a\leq b$.
Thank you so much for answering to my question, but, I understand that the theorem is true iff $\varepsilon $ is small , and how do we know we have a small value of xy in this case?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Thank you so much for answering to my question, but, I understand that the theorem is true iff $\varepsilon $ is small , and how do we know we have a small value of xy in this case?

Suppose on the contrary that $b<a$, then let $\varepsilon = \frac{a - b}{2}$ (which may not be infinitesimal small).
$$
b + \varepsilon = b + \frac{a - b}{2} = \frac{a + b}{2} < \frac{a + a}{2} = a
$$
which is a contradiction.
Therefore, $a\leq b$.
 

Albert

Well-known member
Jan 25, 2013
1,225
x^2+y^2 less then or equal to 1.JPG
 

dwsmith

Well-known member
Feb 1, 2012
1,673

Albert

Well-known member
Jan 25, 2013
1,225
x,y are positive real numbers
$ x^3 +y^3 >0 $
$ so\,\, x\neq y$ (in fact x>y)
x,y be real numbers and meet the given restriction ,so at first I find the range of x and y then prove it will satisfy the inequility
(it is clear x,y can't be any real numbers)
 
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dwsmith

Well-known member
Feb 1, 2012
1,673
x,y are positive real numbers
$ x^3 +y^3 >0 $
$ so\,\, x\neq y$ (in fact x>y)
x,y be real numbers and meet the given restriction ,so at first I find the range of x and y then prove it will satisfy the inequility
(it is clear x,y can't be any real numbers)
Pick up Tom Apostles Real Analysis book. The theorem I stated was for all real numbers not just less than 1
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,689
Here is another method to prove the inequality \(\displaystyle x^2+y^2 \le1\) given \(\displaystyle x^3+y^3\le x-y\) for all real $x$ and $y$. I just saw this solution from another site and immediately wanted to add that solution here....

\(\displaystyle x^3+y^3\le x-y\) tells us \(\displaystyle 0\le y \le x\).

Thus, we have \(\displaystyle x^3\le x\) and this implies \(\displaystyle x \le 1\) and \(\displaystyle 0\le y \le x \le 1\).

In particular, \(\displaystyle x(x+y) \le 1(2) \le 2\;\;\rightarrow\;\;xy(x+y) \le 2y\).

Finally, \(\displaystyle x^3+y^3=(x+y)(x^2-xy+y^2) \le x-y\), so

\(\displaystyle x^2-xy+y^2 \le \frac{x-y}{x+y}\)

\(\displaystyle x^2+y^2 \le \frac{x-y}{x+y}+xy \le \frac{x-y+xy(x+y)}{x+y} \le \frac{x-y+2y}{x+y} \le 1\). (Q.E.D.)