- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,973

**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...

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

Thanks.