Practice – Reasoning and Proof

Case1: Let  $a+b\ge 0$ . Then  $\left|a+b\right|=a+b$ …………..eq.(1)

Also, we know that  $a\le \left|a\right| and b\le \left|b\right|$ , so adding these two inequalities, we get                       $a+b\le \left|a\right|+\left|b\right|$ ,       then from eq.(1), we get

                            $\left|a+b\right|\le \left|a\right|+\left|b\right|$

Case2: Let  $a+b<0$ . Then $\left|a+b\right|=–a–b$ ……………eq.(2)

Also, we know that $–a\le \left|a\right| and –b\le \left|b\right|$ , so adding these two inequalities, we get              $–a–b\le \left|a\right|+\left|b\right|$   ,       then from eq.(2), we get

                          $\left|a+b\right|\le \left|a\right|+\left|b\right|$

[Proof by Contradiction]

Let  $a=\sqrt{x}$ and  $b=\sqrt{y}$ such that  $0\le x\le y$ . Since both  $a$ and  $b$ are non negative and  $x\le y$ , we get  $a^2\le b^2$ .

Let us assume  $b<a$ . Since $b\ge 0$ , we can multiply  $b$ on both sides of  $b<a$ to get       $b^2\le ab$ .

Multiplying $a$ on both sides of  $b<a$ , we get   $ab<a^2$

Putting the above two inequalities together, we get 

$b^2<a^2$ . Which is a contradiction to our hypothesis  $a^2\le b^2$ . So, out assumption that $b<a$ is wrong. Hence  $a\le b$ $\Rightarrow$ $\sqrt{x}<\sqrt{y}$ .

Let  $x$ and  $y$ be real numbers. Now, we know that $0\le \left|x\right|$ and  $0\le \left|y\right|$ , so we can say that                   $0\le 2\left|x\right|\left|y\right|$

Adding $x^2+y^2$ on both sides , we get

                                  $x^2+y^2\le x^2+y^2+2\left|x\right|\left|y\right|$

Also, we know that $x^2={\left|x\right|}^2$   and  $y^2={\left|y\right|}^2$  

$\Rightarrow x^2+y^2\le {\left|x\right|}^2+{\left|y\right|}^2+2\left|x\right|\left|y\right|$

$\Rightarrow x^2+y^2\le {\left(\left|x\right|+\left|y\right|\right)}^2$

$\Rightarrow \sqrt{x^2+y^2}\le \left|x\right|+\left|y\right|$

(a) 0

(b)