Comparing Values

$\dfrac {d}{dx}(\sqrt {x+1}-\sqrt x)$

$=\dfrac {1}{2\sqrt {x+1}}-\dfrac {1}{2\sqrt x}<0$

So, with increase in $x, \sqrt {x+1}-\sqrt x$ decreases. Hence

$\sqrt {99}-\sqrt {98}<\sqrt {20}-\sqrt {19}$

$\implies \boxed {\sqrt {20}+\sqrt {98}>\sqrt {19}+\sqrt {99}}$ .

Define $f(x) = \sqrt x + \sqrt{118-x}$ . Then $f'(x) = \dfrac {df(x)}{dx} = \dfrac 1{2\sqrt x} - \dfrac 1{2\sqrt{118-x}}$ . Putting $f'(x) = 0$ , $\implies x = 118 -x \implies x = 59$ . This means that $f'(x)> 0$ or $f(x)$ is increasing for $x < 58$ . Therefore $f(20) > f(19)$ or $\boxed{\sqrt{20}+\sqrt{98}} > \sqrt{19} + \sqrt{99}$ .

Let $A = \sqrt{19} + \sqrt{99}, B = \sqrt{20} + \sqrt{98}$

• $A^2 = 19 + 99 + 2\sqrt{19\times99} = 118 + 2\sqrt{19\times99}$
• $B^2 = 20+98+ 2\sqrt{20\times98} = 118 + 2\sqrt{20\times98}$

$\therefore$ compare $19 \times 99$ and $20 \times 98$ will get which is larger.

• $19 \times 99 = 19 \times (100-1) = 1881$
• $20 \times 98 = 20 \times (100-2) = 1960 > 1881$

$\therefore B^2 > A^2 \implies B > A$ as both $A$ and $B$ are positive.

$20\cdot98 = 20\cdot100-20\cdot2 = 2000 - 40 > 1900 = 19\cdot100 > 19\cdot99$

$20\cdot98 > 19\cdot99 \implies$

$\sqrt{20\cdot98} > \sqrt{19\cdot99} \implies$

$2\cdot\sqrt{20\cdot98} + 118 > 2\cdot\sqrt{19\cdot99} + 118 \implies$

$(\sqrt{20})^2 + 2\cdot\sqrt{20}\cdot\sqrt{98} + (\sqrt{98})^2 > (\sqrt{19})^2 + 2\cdot\sqrt{19}\cdot\sqrt{99} + (\sqrt{99})^2 \implies$

$(\sqrt{20} + \sqrt{98})^2 > (\sqrt{19} + \sqrt{99})^2 \implies$

$\sqrt{(\sqrt{20} + \sqrt{98})^2} > \sqrt{(\sqrt{19} + \sqrt{99})^2} \implies$

$|\sqrt{20} + \sqrt{98}| > |\sqrt{19} + \sqrt{99}| \implies$

$\boxed{\sqrt{20} + \sqrt{98} > \sqrt{19} + \sqrt{99} }$