# Let x and y be positive real numbers...

• Oct 24th 2008, 01:40 AM
alexmahone
Let x and y be positive real numbers...
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) $x^2 + y^2 < 1$.
• Oct 24th 2008, 08:07 AM
Plato
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?
• Oct 26th 2008, 06:09 AM
simple_life_vu
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