Monday, April 20, 2015

Prove that $x^2+y^2<1$.

Let $x,\,y>0$ be such that $x^3+y^3<x-y$. Prove that $x^2+y^2≤1$.

We have to rate this as a 5 star intriguing Mathematical Olympiad Inequality problem because it is a special problem that is designed for the application of heuristic skills.

I am going to show some sublimely insightful approaches of intelligent people here and as always, I hope you will to learn something awesome today!

>Approach I:

From the given inequality


Before we start, notice that we could glean one really useful information from the given inequality $x-y>x^3+y^3$.

Since both $x$ and $y$ are greater than zero, the sum of two cubes of $x$ and $y$ must also be greater than zero, i.e. $x^3+y^3>0$.

This in turns suggests that [MATH]\color{yellow}\bbox[5px,purple]{x-y>0}[/MATH] since $x-y>x^3+y^3>0$.

We could also algebraically manipulate it such that we could also have


Now, we can safely divide the inequality (*) by the quantity $x-y$ without changing the inequality sign since $x-y>0$ and we obtain:


$1>x^2+y^2$ and we're done.

The heuristic methods that this approach has used are:

a.) the recognition of $x-y>0$,

b.) the recognition of $x^3+y^3>x^3$, and

c.) the introducing of some positive value $y^3+xy(x-y)$ into the expression $x^3-(y^3+xy(x-y))$ and simultaneously the recognition of $x^3>x^3-\text{positive value}$.

Now, it's your turn to attempt at this problem and I can be sure to tell you this problem can be solved differently. 

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