View source for Task:Unnamed inequality

==Statement==
Let {{M|\lambda\in(0,1)\subset\mathbb{R} }} be given. Then for all {{M|t\in\mathbb{R}_{\ge 0} }} we claim:
* {{M|1=t^\lambda \le 1 - \lambda + \lambda t }}
==Proof==
Define: <div style="float:right;margin:0px;margin-left:0.2em;background-color:#F8F8F8;padding:0.2em;border:#A8A8A8 0.1em solid;max-width:25em;">
{| class="wikitable" border="1" style="margin:0px;width:100%;"
|-
! Behaviour:
! {{M|<1}}
! {{M|1==1}}
! {{M|>1}}
|-
! {{M|f'(t)}}
| <center>{{M|<0}}</center>
| <center>{{M|0}}</center>
| <center>{{M|>0}}</center>
|-
! rowspan="2" | {{M|f(t)}}
| <center>{{M|\searrow}}</center>
| <center><div style="margin-top:0.65em;">{{M|\longrightarrow}}</div></center>
| <center>{{M|\nearrow}}</center>
|-
| <center>{{M|>0}}</center>
| <center>{{M|1==0}}</center>
| <center>{{M|>0}}</center>
|-
! colspan="4" | So {{M|f(1)}} is a minimum
|}

<div style="margin:0.5em;">
'''Broughall table for {{M|f(1)}}'''<br/>
Here {{M|d}} is some arbitrarily small positive value.<br/>
'''Reasoning: ''' {{M|t^{\lambda-1} }} for {{M|\lambda\in (0,1)\subset\mathbb{R} }} means {{M|\lambda-1\in(-1,0)\subset\mathbb{R} }}. Thus {{M|1=(1\pm d)^{\lambda-1}=\frac{1}{(1\pm d)^{1-\lambda} } }} and now {{M|1-\lambda\in(0,1)}}. 
# Thus {{M|(1-d)^{1-\lambda} < 1}}
#* so {{M|\frac{1}{(1-d)^{1-\lambda} } > 1}} 
#* so {{M|1-\frac{1}{(1-d)^{1-\lambda} } <0}} 
#* and finally {{M|1=\lambda(1-\frac{1}{(1-d)^{1-\lambda} }) <0}}
#** As {{M|\lambda\in(0,1)}} (i.e. "positive")
#* Thus {{M|1=\lambda(1-(1-d)^{\lambda-1})<0}}
# Thus {{M|(1+d)^{1-\lambda} > 1}}
#* so {{M|\frac{1}{(1+d)^{1-\lambda} } < 1}} 
#* so {{M|1-\frac{1}{(1+d)^{1-\lambda} } >0}} 
#* and finally {{M|1=\lambda(1-\frac{1}{(1+d)^{1-\lambda} }) >0}}
#** As {{M|\lambda\in(0,1)}}
#* Thus {{M|1=\lambda(1-(1+d)^{\lambda-1})>0}}
</div>
</div>
* {{M|f:\mathbb{R}_{\ge 0}\rightarrow\mathbb{R} }} by {{M|f:t\mapsto 1-\lambda+\lambda t-t^\lambda }}, so {{M|1=f(t):=1-\lambda+\lambda t - t^\lambda }}. Then:
** {{M|1=f':t\mapsto \lambda - \lambda t^{\lambda-1} }} by [[differentiation]] (for {{M|f':\mathbb{R}_{>0}\rightarrow\mathbb{R} }} - we'd have to use a [[limit-from-the-right]] to find {{M|f'(0)}} if it exists.
*** We wish to find a [[turning point]] of {{M|1=f'(t)=\lambda - \lambda t^{\lambda-1}=\lambda(1-t^{\lambda-1})}}, that is {{M|1=f'(a)=0}} for some {{M|a}}
**** Suppose {{M|1=\lambda - \lambda t^{\lambda-1} = 0}}, then {{M|1=\lambda - \lambda t^{\lambda-1}=\lambda(1-t^{\lambda-1})=0}} and clearly:
***** If {{M|1=t=1}} then {{M|1=1-t^{\lambda-1}=1-1=0}} and we're done (there is no other [[root (function)|root]])
*** Thus {{M|1}} is a turning point of {{M|f}} (that is {{M|1=f'(1)=0}})
*** Next we must find out what {{M|f}} is doing. We can use a [[Broughall table]] (shown right)
**** Or we can use the second derivative:
***** {{M|1=f' '(t)=-\lambda(\lambda-1)t^{\lambda-2} }}
****** The [[quadratic curve]] {{M|\lambda(\lambda-1)}} has roots {{M|0}} and {{M|1}} and tends towards {{M|+\infty}} as tends to {{M|\pm\infty}}, thus is standard {{M|\cup}} shaped.
****** The quadratic curve {{M|-\lambda(\lambda-1)}} is the same but flipped about the {{M|x}}-axis. Thus a {{M|\cap}} shape.
****** As {{M|\lambda\in(0,1)}} and this is between the roots, we see {{M|-\lambda(\lambda-1)<0}} over the relevant domain. 
****** {{M|t > 0}} so {{M|t^\text{anything real} > 0}} (remember we have to do {{M|1=t=0}} separately, with a [[directed limit]])
***** So {{M|-\lambda(\lambda-1)t^{\lambda-2} }} is a {{M|\cap}}-shape with respect to {{M|\lambda}} and {{M|>0}} always on the relevant {{M|\lambda}}-domain.
***** We conclude {{M|f' '(1) > 0}} (we can use this in the Broughall table, knowing {{M|1=f'(1)=0}} and is increasing at {{M|1}} means {{M|f'(\text{slightly after }1)>0}} and as it must increase to zero {{M|1=f'(\text{slightly before }1)<0}}, as shown
***** Thus {{M|f(1)}} is a minimum
*** We have shown {{M|f}} is at a global minimum for all relevant {{M|\lambda}} and all {{M|t}} except {{M|1=t=0}}
**** We evaluate {{M|f(1)}} to get {{M|1=f(1)=0}}. Thus {{M|f(t)>0}} for {{M|t\in\mathbb{R}_{>0} }} (but not {{M|1=t=0}} yet!)
*** We evaluate {{M|f(0)}} to get {{M|1-\lambda}} (as {{M|1=t^\lambda=0^\lambda=0}} as {{M|\lambda\in(0,1)}} and {{M|0^x}} is only undefined for {{M|1=x=0}})
**** We observe that as {{M|\lambda\in(0,1)}} that {{M|1-\lambda\in(0,1)}} also, thus {{M|1=f(0)=1-\lambda>0}} as required.
** We have seen that {{M|1=f(t)\ge 0}} for all {{M|1=t\in\mathbb{R}_{\ge 0} }}
*** Thus {{M|1=1-\lambda+\lambda t-t^\lambda\ge 0}} (we will no longer explicitly mention the domain of {{M|t}} or {{M|\lambda}})
*** So {{M|1=1-\lambda+\lambda t \ge t^\lambda }}
**** As required
* This completes the proof.