# Prove that $2^{n-1}(a^n+b^n)\geq(a+b)^n$ [duplicate]

I'm trying to prove for algorithm-purposes that given $$a,b,n$$ positive integers: $$2^{n-1}(a^n+b^n)\geq(a+b)^n$$ I tried by induction, and got the step as follows: $$2^{n}(a^{n+1}+b^{n+1})\geq^?(a+b)^{n+1}$$ I tried using the binomial expansion $$(a+b)^n=\sum^n_{k=0} {{n}\choose{k}}a^kb^{n-k}$$ and then excluding the last element $$(a+b)^{n+1}=\sum^{n+1}_{k=0} {{n+1}\choose{k}}a^kb^{n-k+1}=\sum^{n}_{k=0} {{n+1}\choose{k}}a^kb^{n-k}b+{{n+1}\choose{n+1}}a^{n+1}b^0$$$$=\sum^{n}_{k=0} (n+1){{n}\choose{k}}a^kb^{n-k}b+{{n+1}\choose{n+1}}a^{n+1}b^0=[(n+1)b]\sum^{n}_{k=0}{{n}\choose{k}}a^kb^{n-k}+{{n+1}\choose{n+1}}a^{n+1}b^0$$$$=[(n+1)b](a+b)^n+a^{n+1}\leq[(n+1)b]\times2^{n-1}(a^n+b^n)+a^{n+1}$$ Assuming that everything is correct so far, I don't how to proceed from there to get $$\leq 2^n(a^{a+1}+b^{n+1})$$

My second attempt was by doing the step as follows: $$2^{n-1}(a^n+b^n)\geq(a+b)^n \setminus\cdot(a+b)$$ $$2^{n-1}(a^n+b^n)(a+b)\geq(a+b)^{n+1}$$ $$2^{n-1}(a^{n+1}+b^{n+1}+a^nb+b^na)\geq(a+b)^{n+1}$$ Now I don't know how to eliminate $$a^nb+b^na$$, and to proceed to $$2^n$$

Is there another way to prove this? Or any hints to continue my step?

From the rearrangement inequality, we have the following: $$a^n+b^n\geq a^i\cdot b^{n-i}+a^{n-i}\cdot b^i$$
$$(a+b)^n=\sum_{i=0}^{n}\binom{n}{i}a^ib^{n-i}\leq\frac{a^n+b^n}{2}\sum_{i=1}^{n}\binom{n}{i}$$