Prove by induction that for every natural number $n$ is worth $2^n + 3^n ≤ 5^n$

Question

I started the exercise in the following way: Let $p(n): 2^n + 3^n ≤ 5^n$ be a propositional function. $p(1): 2^1 + 3^1 = 5 ≤ 5^1 = 5$ it's true. Now, let $k$ be a natural number. I assume that $p(k)$ is true and I want to know if $p(k+1)$ it's true having $p(k)$ as hypothesis but I can not do it. I understand the mechanics of induction exercises and I have been able to do others but with this I can't prove $p(k+1)$ using $p(k)$.

Answer 1

$$2^{n+1} + 3^{n+1} =$$

$$ 2(2^n)+3(3^n) < 5(2^n)+5(3^n)=$$

$$5(2^n+3^n)\le 5(5^n)=5^{n+1}$$

Answer 2

$$2^{n+1}=2 \cdot 2^n\le 5 \cdot 2^n $$

$$3^{n+1}=3 \cdot 3^n\le 5 \cdot 3^n $$

thus

$$2^{n+1}+3^{n+1}\le 5 (2^n+3^n)\le 5^{n+1} $$

Answer 3

Not by induction, but the binomial theorem is better I think because it shows you why it is true:

$$5^n = (2+3)^n = \color{red}{2^n} + {n \choose 1}2^{n-1} \cdot3 + \cdots + {n \choose n-1} 2 \cdot 3^{n-1} + \color{red}{3^n} \ge 2^n+3^n$$

So many things in math are about doing two things in two different orders and asking what happens. Here you either add $2$ and $3$ and rais to a power, or you raise to a power first and then add. As seen from the binomial expansion, the second way gives you a much smaller number.

Answer 4

Alt. hint: prove (by induction, if you must) that $\,a^n \le a\,$ for $\,a \in [0,1]\,$ and $\,n \ge 1\,$, then:

$$\,\left(\dfrac{2}{5}\right)^n+\left(\dfrac{3}{5}\right)^n \,\le\, \dfrac{2}{5}+ \dfrac{3}{5} \,=\, 1\,$$