Let x and y be positive real numbers...

alexmahone · #1 $Old$ 10-24-2008, 03:40 AM

Let x and y be positive real numbers such that $y^3 + y \le x - x^3$ . Prove that
(a) $y < x < 1$ ; and
(b) $x^2 + y^2 < 1$ .

Plato · #2 $Old$ 10-24-2008, 10:07 AM

From the given, $x > 0\,\& \,y > 0$ , we get
$\begin{array}{rcl} {y^3 + y \leqslant x - x^3 } & \Leftrightarrow & {y\left( {y^2 + 1} \right) \leqslant x\left( {1 - x^2 } \right)} \\ {} & \Leftrightarrow & {0 < \frac{y}{x} \leqslant \frac{{1 - x^2 }}{{y^2 + 1}}} \\ {} & \Leftrightarrow & {0 < 1 - x^2 } \\ {} & \Leftrightarrow & {0 < x < 1} \\ \end{array}$ .

From the above we get
$y \geqslant x \Rightarrow \quad y^3 + y \geqslant x + x^3 \Rightarrow \quad y^3 + y > x - x^3$ .

Do you see ho to finish?

simple_life_vu · #3 $Old$ 10-26-2008, 08:09 AM

a) $y^3 + y$ is positive because y>0, hence $x - x^3 > 0 \Rightarrow x ( 1 - x^2 ) > 0 \Rightarrow 1 - x^2 > 0 \Rightarrow x < 1$
$y^3 + y \le x - x^3 \Rightarrow y^3 + x^3 \le x - y$
Because LHS is greater than 0 ( x and y are positive real numbers ) hence
x-y > 0 and x>y

b) x,y< 1
Hence $x^2 > x^4$ (1)and
$y^2 > y^4$ (2)

Take away (2) from (1) and factorise $\Rightarrow x^2 - y^2 > ( x^2 - y^2 ) ( x^2 + y^2 )$

Now x > y hence $x^2 - y^2 > 0$ and we can divide both side by $x^2 - y^2 > 0$ and get the desired inequality