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Chapter 2: Problem 140

Show that if \(x\) and \(y\) are positive numbers, then $$ \sqrt{x+y}<\sqrt{x}+\sqrt{y} $$ [In particular, if \(x\) and \(y\) are positive numbers, then \(\sqrt{x+y} \neq \sqrt{x}+\sqrt{y} .]\)

Short Answer

Expert verified
To show that \(\sqrt{x+y} < \sqrt{x} + \sqrt{y}\) for positive numbers \(x\) and \(y\), we squared both sides of the inequality, simplified, and rearranged to obtain \(0 < \sqrt{xy}\). Since this inequality is true for positive \(x\) and \(y\), we concluded that \(\sqrt{x+y} < \sqrt{x} + \sqrt{y}\) and in particular, \(\sqrt{x+y} \neq \sqrt{x} + \sqrt{y} .\)

Step by step solution

01

Square both sides of the inequality

To compare the expressions \(\sqrt{x+y}\) and \(\sqrt{x} + \sqrt{y}\), start by squaring both sides of the inequality: \[ (\sqrt{x+y})^2 < (\sqrt{x} + \sqrt{y})^2 \]
02

Simplify both sides of the inequality

Next, simplify both sides of the inequality. Recall that squaring a square root cancels the square root, so we obtain: \[ x + y < (\sqrt{x} + \sqrt{y})^2 \] Expand the right side of the inequality using the binomial formula: \((a+b)^2 = a^2 + 2ab + b^2\). In this case, \(a = \sqrt{x}\) and \(b = \sqrt{y}\): \[ x + y < (\sqrt{x})^2 + 2(\sqrt{x})(\sqrt{y}) + (\sqrt{y})^2 \] Simplify the right side further: \[ x + y < x + 2\sqrt{xy} + y \]
03

Rearrange the inequality

Subtract \(x\) and \(y\) from both sides to isolate the term involving both \(x\) and \(y\) and simplify further: \[ 0 < 2\sqrt{xy} \] Now we can divide both sides of the inequality by 2: \[ 0 < \sqrt{xy} \]
04

Discuss the result

As we can see, the inequality \(0 < \sqrt{xy}\) is true for positive numbers \(x\) and \(y\), since the square root of a positive number is positive. This means that we have showed that: \[ \sqrt{x+y} < \sqrt{x} + \sqrt{y} \] for all positive numbers \(x\) and \(y\). In particular, the given inequality implies that \(\sqrt{x+y} \neq \sqrt{x} + \sqrt{y} .\)

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