inequality proof

symbian · **Joined:** Thu, 23 Sep 2010 10:34:44 UTC **Posts:** 32

With reference to Bartle and Sherbert Introduction to Real Analysis Pg 28 Example 2.1.13

May I know how to proof this?

$Let $a \geq 0$ and $b\geq 0$. Prove that $a \leq b \Leftrightarrow a^2 \leq b^2 \Leftrightarrow \sqrt{a} \leq \sqrt{b}$

Shadow · **Posted:** Sat, 21 May 2011 17:01:28 UTC

symbian wrote:

With reference to Bartle and Sherbert Introduction to Real Analysis Pg 28 Example 2.1.13

May I know how to proof this?

$Let $a \geq 0$ and $b\geq 0$. Prove that $a \leq b \Leftrightarrow a^2 \leq b^2 \Leftrightarrow \sqrt{a} \leq \sqrt{b}$

$a\cdot a\le a\cdot b\le b\cdot b$ by positivity of a and b.

Similarly the square root follows from the first one with new a being the old a^2

and new b being old b^2

.

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

Let x and y be positive numbers. If x ≤ y, then √x ≤ √y.

Proof. Suppose x ≤ y.
Subtracting y from both sides gives x - y ≤ 0 ⇔ (√x)^2 - (√y)^2 ≤ 0.
Factor this to get (√x - √y)(√x + √y) ≤ 0.
Dividing both sides by the positive number √x + √y produces √x - √y ≤ 0.
Adding √y to both sides gives √x ≤ √y.

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inequality proof

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