Challenge of Square Root Problem

[anemone]

Show that $\sqrt{99}-\sqrt{98}+\sqrt{97}-\sqrt{96}+\cdots-\sqrt{4}+\sqrt{3}-\sqrt{2}+\sqrt{1}> 5$.

 

[Opalg]

Show that $\sqrt{99}-\sqrt{98}+\sqrt{97}-\sqrt{96}+\cdots-\sqrt{4}+\sqrt{3}-\sqrt{2}+\sqrt{1}> 5$.

[sp]
First, $\sqrt{2n+1} - \sqrt{2n} = \dfrac1{\sqrt{2n+1} + \sqrt{2n}} > \dfrac1{2\sqrt{2n+1}}.$ Therefore $$\sqrt{99}-\sqrt{98}+\sqrt{97}-\sqrt{96}+\ldots-\sqrt{4}+\sqrt{3}-\sqrt{2}+\sqrt{1} > \frac12\left(\frac1{\sqrt{99}} + \frac1{\sqrt{97}} + \frac1{\sqrt{95}} + \ldots + \frac1{\sqrt3}\right) + 1.$$ Next, $$\frac1{\sqrt3} >\frac1{\sqrt4} = \frac12,$$ $$\frac1{\sqrt5} + \frac1{\sqrt7} + \frac1{\sqrt9} > \frac3{\sqrt9} = \frac33,$$ $$\frac1{\sqrt{11}} + \frac1{\sqrt{13}} + \frac1{\sqrt{15}} > \frac3{\sqrt16} = \frac34,$$ and so on, up to $$\frac1{\sqrt{83}} + \frac1{\sqrt{85}} +\frac1{\sqrt{87}} +\ldots + \frac1{\sqrt{99}} > \frac9{\sqrt{100}} = \frac9{10}.$$ Putting those all together, we get $$\frac1{\sqrt{99}} + \frac1{\sqrt{97}} + \frac1{\sqrt{95}} + \ldots + \frac1{\sqrt3} > \frac9{10} + \frac99 + \frac78 + \frac77 + \frac56 + \frac55 + \frac34 + \frac33 + \frac12 = 8\tfrac{103}{120} > 8.$$ Finally, $$\begin{aligned}\sqrt{99}-\sqrt{98}+\sqrt{97}-\sqrt{96}+\ldots-\sqrt{4}+\sqrt{3}-\sqrt{2}+\sqrt{1} &> \frac12\left(\frac1{\sqrt{99}} + \frac1{\sqrt{97}} + \frac1{\sqrt{95}} + \ldots + \frac1{\sqrt3}\right) + 1 \\ &> \frac82 + 1 = 5.\end{aligned}$$
[/sp]

 

[anemone]

[sp]
First, $\sqrt{2n+1} - \sqrt{2n} = \dfrac1{\sqrt{2n+1} + \sqrt{2n}} > \dfrac1{2\sqrt{2n+1}}.$ Therefore $$\sqrt{99}-\sqrt{98}+\sqrt{97}-\sqrt{96}+\ldots-\sqrt{4}+\sqrt{3}-\sqrt{2}+\sqrt{1} > \frac12\left(\frac1{\sqrt{99}} + \frac1{\sqrt{97}} + \frac1{\sqrt{95}} + \ldots + \frac1{\sqrt3}\right) + 1.$$ Next, $$\frac1{\sqrt3} >\frac1{\sqrt4} = \frac12,$$ $$\frac1{\sqrt5} + \frac1{\sqrt7} + \frac1{\sqrt9} > \frac3{\sqrt9} = \frac33,$$ $$\frac1{\sqrt{11}} + \frac1{\sqrt{13}} + \frac1{\sqrt{15}} > \frac3{\sqrt16} = \frac34,$$ and so on, up to $$\frac1{\sqrt{83}} + \frac1{\sqrt{85}} +\frac1{\sqrt{87}} +\ldots + \frac1{\sqrt{99}} > \frac9{\sqrt{100}} = \frac9{10}.$$ Putting those all together, we get $$\frac1{\sqrt{99}} + \frac1{\sqrt{97}} + \frac1{\sqrt{95}} + \ldots + \frac1{\sqrt3} > \frac9{10} + \frac99 + \frac78 + \frac77 + \frac56 + \frac55 + \frac34 + \frac33 + \frac12 = 8\tfrac{103}{120} > 8.$$ Finally, $$\begin{aligned}\sqrt{99}-\sqrt{98}+\sqrt{97}-\sqrt{96}+\ldots-\sqrt{4}+\sqrt{3}-\sqrt{2}+\sqrt{1} &> \frac12\left(\frac1{\sqrt{99}} + \frac1{\sqrt{97}} + \frac1{\sqrt{95}} + \ldots + \frac1{\sqrt3}\right) + 1 \\ &> \frac82 + 1 = 5.\end{aligned}$$
[/sp]

Very well done, Opalg! (Smile)

Solution of other:

Spoiler

Let

$X=\sqrt{100}-\sqrt{99}+\sqrt{98}-\sqrt{97}-\sqrt{96}+\cdots-\sqrt{4}+\sqrt{3}$, and

$Y+1=\sqrt{99}-\sqrt{98}+\sqrt{97}-\sqrt{96}+\cdots-\sqrt{4}+\sqrt{3}-\sqrt{2}+\sqrt{1}$.

Adding up the equations yields $X+Y=\sqrt{100}-\sqrt{2}$.

Note that from AM-GM inequality, we have the following inequalities that are true:

$100+98>2\sqrt{100}\sqrt{98}$

$4(99)>100+2\sqrt{100}\sqrt{98}+98$

$2\sqrt{99}>\sqrt{100}+\sqrt{98}$

\(\displaystyle \color{yellow}\bbox[5px,purple]{\sqrt{99}-\sqrt{98}>\sqrt{100}-\sqrt{99}}\),

\(\displaystyle \color{yellow}\bbox[5px,green]{\sqrt{97}-\sqrt{96}>\sqrt{98}-\sqrt{97}}\), and so on and so forth, we know $Y>X$ must hold.

Therefore, $2Y>X+Y=\sqrt{100}-\sqrt{2}$ (from $X+Y=\sqrt{100}-\sqrt{2}$), and so

$Y+1>\dfrac{\sqrt{100}-\sqrt{2}}{2}+1=5+\left(1-\dfrac{\sqrt{2}}{2}\right)>5$ and we're hence done.