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Problem Of The Week #391 Nov 7th, 2019

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
No one answered last week's POTW.

Below is a suggested solution from other:
If both $x$ and $y$ $\le0$, then the inequality is obvious.

If both $x$ and $y$ $>1$, then they do not satisfy the condition. So either $x\ge 1 \ge y$ or $y\ge 1 \ge x$ is true.

We show first that $x^3+y^3\le x^2+y^2$.

In the first case, we have

$x^2+y^2(1-y)\ge x^2+y^3(1-y)\ge x^3+y^4-y^4=x^3$

So $x^2+y^2\ge x^3+y^3$ as required.

In the second case, we have

$x^2\ge x^3+y^3(y-1)\ge x^3+y^2(y-1)$

So again $x^2+y^2\ge x^3+y^3$

Now by Cauchy-Schwartz inequality,

$x^2+y^2=x^{\frac{3}{2}}x^{\frac{1}{2}}+y^{\frac{3}{2}}y^{\frac{1}{2}}\le \sqrt{x^3+y^3}\sqrt{x+y}$

But $x^2+y^2\ge x^3+y3$ so $x^2+y2\le x+y$

Since $2xy\le x^2+y2$ so $(x+y)^2\le 2(x^2+y^2)$

Since we have proved $2(x^2+y^2)\le 2(x+y)$ we therefore obtain

$x+y\le 2$.

Thus we have

$x^3+y^3\le x^2+y^2\le x+y \le 2$.

 
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