# Cubed Roots Proof

• Mar 1st 2011, 03:10 PM
Berge
Cubed Roots Proof
(Headbang)...How can I show that If 0<x<y then :

$\displaystyle \noindent$$\pmb{\sqrt[3]{y-x}> \sqrt[3]{y}-\sqrt[3]{x}}$$$
• Mar 1st 2011, 06:18 PM
tonio
Quote:

Originally Posted by Berge
(Headbang)...How can I show that If 0<x<y then :

$\displaystyle \noindent$$\pmb{\sqrt[3]{y-x}> \sqrt[3]{y}-\sqrt[3]{x}}$$$

Raising to the 3rd power both sides:

$\displaystyle \sqrt[3]{y-x}>\sqrt[3]{y}-\sqrt[3]{x}\Longleftrightarrow y-x>y-3y^{2/3}x^{1/3}+3y^{1/3}x^{2/3}-x\Longleftrightarrow$

$\displaystyle \Longleftrightarrow 3\sqrt[3]{xy}(\sqrt[3]{x}-\sqrt[3]{y})<0$ , and since this last inequality is trivial we're done.

Tonio
• Mar 2nd 2011, 02:50 AM
HallsofIvy
What Tonio did is an example of what is called in some texts "synthetic proof". Notice that he started from the thing he wanted to prove and manipulated it to a statement that is "trivially" true. Strictly speaking, you are not allowed to assume what you want to prove but here everything is "reversible" (note the double ended arrows). The "real" proof would start from $\displaystyle 3\sqrt[3]{xy}\left(\sqrt[3]{x}-\sqrt[3]{y}\right)< 0$, which is obviously true because x< y, and work backwards to $\displaystyle \sqrt[3]{y- x}> \sqrt[3]{y}- \sqrt[3]{x}$.