Prove the inequality: $1.6^{n-2}+1.6^{n-2} \ge 1.6^{n-1}$

Question

I need a more simplistic proof that this inequality holds for all $n > 2$:

$$1.6^{n-2}+1.6^{n-2} \ge 1.6^{n-1}$$

I think I could manage it using derivates to show that the function is always decreasing but I feel there's a simpler way.

Answer 1

Yes, here is a simpler way: \begin{align*} 1.6^{n-2} + 1.6^{n-2} &= 2 \left( 1.6^{n-2} \right) \\ &> 1.6 \left( 1.6^{n-2} \right) \\ &= 1.6^{n-1}. \end{align*}

Answer 2

$$1.6^{n-1} = 1.6\cdot1.6^{n-2}$$

On the other hand,

$$1.6^{n-2} + 1.6^{n-2} = 2\cdot1.6^{n-2}$$

Can you take it from here?

Answer 3

$$\frac{8^{n-2}}{5^{n-2}}+\frac{8^{n-2}}{5^{n-2}}=2\frac{8^{n-2}}{5^{n-2}}\ge\frac{8^{n-1}}{5^{n-1}}\iff2\ge\frac85\;\;\;\color{green}\checkmark$$

Answer 4

$${1.6^{n-2}}+1.6^{n-2}=1.6^{n-2}(2)>1.6^{n-2}(1.6)=1.6^{n-1}$$