Why is 0≥1? (1 Viewer)

HeroicPandas · Apr 27, 2014

Alright, fresh start... sorry

Beginning from the use of induction hypothesis

$\geq \frac{2}{3}k\sqrt{k} + \sqrt{k+1}$

$= \frac{2}{3}\sqrt{k+1} \left ( \frac{k\sqrt{k}}{\sqrt{k+1}} +\frac{3}{2}\right )$

Now we need to prove that (time for a discovery)

$\frac{k\sqrt{k}}{\sqrt{k+1}} +\frac{3}{2}\geq k+1$

Subtract 1.5 on both sides

$\frac{k\sqrt{k}}{\sqrt{k+1}} \geq k-\frac{1}{2}$

Multiply both sides by 2sqrt(k+1)

$2k\sqrt{k}\geq 2k\sqrt{k+1} - \sqrt{k+1}$

$2k(\sqrt{k}-\sqrt{k+1}) + \sqrt{k+1} \geq 0$

Edits:

Actually.. as k>0 and sqrt(k+1)>0, then sqrt(k) - sqrt(k+1)> 0 for LHS >0, so k>k+1, 0>1..................

When does equality occur?

SpiralFlex · Apr 27, 2014

HeroicPandas said:
Alright, fresh start... sorry

Beginning from the use of induction hypothesis

$\geq \frac{2}{3}k\sqrt{k} + \sqrt{k+1}$

$= \frac{2}{3}\sqrt{k+1} \left ( \frac{k\sqrt{k}}{\sqrt{k+1}} +\frac{3}{2}\right )$

Now we need to prove that (time for a discovery)

$\frac{k\sqrt{k}}{\sqrt{k+1}} +\frac{3}{2}\geq k+1$

Subtract 1 on both sides

$\frac{k\sqrt{k}}{\sqrt{k+1}} \geq k-\frac{1}{2}$

Multiply both sides by 2sqrt(k+1)

$2k\sqrt{k}\geq 2k\sqrt{k+1} - \sqrt{k+1}$

$2k(\sqrt{k}-\sqrt{k+1}) + \sqrt{k+1} \geq 0$

Edits:

Actually.. as k>0 and sqrt(k+1)>0, then sqrt(k) - sqrt(k+1)> 0 for LHS >0, so k>k+1, 0>1..................

When does equality occur?

Counter example k = 1. Read my spoiler.

HeroicPandas · Apr 27, 2014

SpiralFlex said:
Counter example k = 1. Read my spoiler.

Ahhh subtracting 1 (meant to be 1.5) on both sides destroyed my solution

About your solution, the last line, when does equality occur for $k \in \mathbb{Z}^+$ ?

Why is 0≥1? (1 Viewer)

HeroicPandas

Heroic!

SpiralFlex

Well-Known Member

HeroicPandas

Heroic!

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