MHB How Can We Prove the Modified Bernoulli Inequality?

[evinda]

Hello! (Wave)

Using induction, I have showed the Bernoulli inequality, i.e. that if $a \geq -1$ and $n \in \mathbb{N}$ then $1+na \leq (1+a)^n$. Now I want to show that if $a \geq -1$ and $n \in \mathbb{N}$ the $1+\frac{1}{n}a \geq (1+a)^{\frac{1}{n}}$. How could we show this? Could we use somehow the Bernoulli inequality? (Thinking)

 

[I like Serena]

Hey evinda! (Smile)

Suppose we substitute $a=\frac 1n \tilde a$ in Bernoulli's inequality. Would that bring us closer to what we want to prove? (Wondering)

 

[evinda]

Hey evinda! (Smile)

Suppose we substitute $a=\frac 1n \tilde a$ in Bernoulli's inequality. Would that bring us closer to what we want to prove? (Wondering)

Then we get that $1+ \tilde a \leq \left( 1+ \frac{1}{n} \tilde a\right)^n \Rightarrow 1+\frac{1}{n} \tilde a \geq (1+ \tilde a)^{\frac{1}{n}}$, right?

But this holds for $\tilde a \geq -n$. We want that it holds for $\tilde a \geq -1$. (Worried) What could we do? (Thinking)

 

[I like Serena]

Doesn't the first imply the second?
We actually found a 'stronger' inequality than we need. (Thinking)

 

[evinda]

Doesn't first imply the second?
We actually found a 'stronger' inequality than we need. (Thinking)

Ah because if the inequality holds for values greater than $-n$, we directly have that it holds for values greater than $-1$, right? (Blush)

 

[I like Serena]

Yep. (Nod)

 

[evinda]

Yep. ( Nod)

I see... Thanks a lot! (Smirk)