# Is proof valid?

• Mar 9th 2006, 03:46 AM
DenMac21
Is proof valid?
I need to prove that inequality is true:

$\displaystyle \sqrt a + \sqrt b \le \sqrt {\frac{{a^2 }}{b}} + \sqrt {\frac{{b^2 }}{a}}$ for $\displaystyle a,b>0$

Transforming it we get:
$\displaystyle \sqrt a + \sqrt b \le \frac{a}{{\sqrt b }} + \frac{b}{{\sqrt a }}$
$\displaystyle \frac{{b - a}}{{\sqrt b }} \le \frac{{b - a}}{{\sqrt a }}$
$\displaystyle \frac{{\sqrt a }}{{\sqrt b }} \le \frac{{b - a}}{{b - a}}$
$\displaystyle \frac{{\sqrt a }}{{\sqrt b }} \le 1 \Leftrightarrow \sqrt a \le \sqrt b \Leftrightarrow a \le b$

Now, from $\displaystyle \frac{{b - a}}{{\sqrt b }} \le \frac{{b - a}}{{\sqrt a }}$ we get $\displaystyle \sqrt a (b - a) - \sqrt b (b - a) \le 0 \Leftrightarrow (b - a)(\sqrt a - \sqrt b ) \le 0$ which is true because from first transformation we get that $\displaystyle a \le b$ and $\displaystyle \sqrt a \le \sqrt b$.

Is this proof valid?
• Mar 9th 2006, 04:23 AM
topsquark
Quote:

Originally Posted by DenMac21
I need to prove that inequality is true:

$\displaystyle \sqrt a + \sqrt b \le \sqrt {\frac{{a^2 }}{b}} + \sqrt {\frac{{b^2 }}{a}}$ for $\displaystyle a,b>0$

Transforming it we get:
[
$\displaystyle \frac{{b - a}}{{\sqrt b }} \le \frac{{b - a}}{{\sqrt a }}$
$\displaystyle \frac{{\sqrt a }}{{\sqrt b }} \le \frac{{b - a}}{{b - a}}$

This statement is only true if b - a > 0. An example of how this fails: Set a=4 and b=1 in the original expression. This expression is correct. Your final result, though, predicts that 4 < 1. The reason is in the line above. If b - a is negative, the inequality switches direction.

-Dan
• Mar 9th 2006, 04:26 AM
CaptainBlack
Something is wrong.

You wish to prove that:

$\displaystyle \sqrt a + \sqrt b \le \sqrt {\frac{{a^2 }}{b}} + \sqrt {\frac{{b^2 }}{a}}$

is an identity for $\displaystyle a,b>0$.

You assume it true and deduce that:

$\displaystyle a \le b$.

Hence if $\displaystyle b < a$ you have a contradiction.

RonL