Task:Unnamed inequality

Statement

Let $\lambda\in(0,1)\subset\mathbb{R}$ be given. Then for all $t\in\mathbb{R}_{\ge 0}$ we claim:

• $t^\lambda \le 1 - \lambda + \lambda t$

Proof

Define:

Behaviour:	$<1$	$=1$	$>1$
$f'(t)$	$<0$	$0$	$>0$
$f(t)$	$\searrow$	$\longrightarrow$	$\nearrow$
	$>0$	$=0$	$>0$
So $f(1)$ is a minimum

Broughall table for $f(1)$
Here $d$ is some arbitrarily small positive value.
Reasoning: $t^{\lambda-1}$ for $\lambda\in (0,1)\subset\mathbb{R}$ means $\lambda-1\in(-1,0)\subset\mathbb{R}$. Thus $(1\pm d)^{\lambda-1}=\frac{1}{(1\pm d)^{1-\lambda} }$ and now $1-\lambda\in(0,1)$.

1. Thus $(1-d)^{1-\lambda} < 1$
   • so $\frac{1}{(1-d)^{1-\lambda} } > 1$
   • so $1-\frac{1}{(1-d)^{1-\lambda} } <0$
   • and finally $\lambda(1-\frac{1}{(1-d)^{1-\lambda} }) <0$
     • As $\lambda\in(0,1)$ (i.e. "positive")
   • Thus $\lambda(1-(1-d)^{\lambda-1})<0$
2. Thus $(1+d)^{1-\lambda} > 1$
   • so $\frac{1}{(1+d)^{1-\lambda} } < 1$
   • so $1-\frac{1}{(1+d)^{1-\lambda} } >0$
   • and finally $\lambda(1-\frac{1}{(1+d)^{1-\lambda} }) >0$
     • As $\lambda\in(0,1)$
   • Thus $\lambda(1-(1+d)^{\lambda-1})>0$

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