Prove that $a_n \gt b_n$ $\forall $ $n \ge 6$

Question

Given that $$a_n =\left(1^2+2^2+3^2+\cdots+n^2\right)^n$$ and

$$b_n=n^n \times (n!)^2$$ Then prove that $a_n \gt b_n$ $\forall $ $n \ge 6$

My attempt:

I tried using induction, but I could not prove the basic step i.e to prove that the proposition $P(6)$ is True.

$$a_6=(273)^6$$

$$b_6=(6^6)(720)^2$$

But how can we prove $a_6 \gt b_6$ without calculator?

Answer 1

Because by AM-GM we obtain: $$\sqrt{\frac{1^2+2^2+...+n^2}{n}}\geq\sqrt[n]{n!}.$$

Answer 2

I'm not sure where $273$ came from; $1^2+2^2+3^2+4^2+5^2+6^2=1+4+9+16+25+36=91$

I'd go about proving $a_6>b_6$ as follows (without even doing the above computation). First, note that we have a lot of sixth powers floating around, except the $720$ is squared. Let's turn that into a sixth power by approximating $720$ by an easy cube: $$720 < 1000 = 10^3$$ $$720^2 < (10^3)^2 = 10^6$$ Thus $$b_6= 6^6720^2 < 6^610^6=60^6$$ Now let's see if we can get $a_6>60^6$, i.e., is $1^2+ \dots + 6^2>60$? The greatest term is $6^2=36$, not enough yet; but once we take two terms, $6^2+5^2=36+25=61>60$. Thus $$a_6 > (6^2+5^2)^6 = 61^6 > 60^6 > b_6$$

Answer 3

The Arithmetic-Geometric Mean Inequality handles things for all $n\gt1$ without the need for induction (beyond the fact that AGM itself is typically proved by induction):

$$(1^2+2^2+\cdot+n^2)^n\gt n^n(n!)^2\iff{1^2+2^2+\cdots+n^2\over n}\gt\sqrt[n]{1^2\cdot2^2\cdots(n-1)^2\cdot n^2}$$

with strict inequality guaranteed by the fact that $1^2\not=2^2$ (i.e., the terms being averaged are not all equal).

I'm not sure why the problem would ask for a proof that only starts at $n=6$; it looks like a red herring. But if you do want an easy proof for that case, note that

$$6^6(6!)^2=6^6\cdot720^2\lt6^6\cdot729^2=6^6\cdot(9^3)^2=54^6$$

and $54\lt36+25=6^2+5^2\lt1^2+2^2+3^2+4^2+5^2+6^2$.

Answer 4

It is not hard first to prove that for $n\ge 6$ it is: $$1^2+2^2+3^2+\cdots+n^2 \ge n^3 $$ then of course we will have$$\left(1^2+2^2+3^2+\cdots+n^2\right)^n\ge n^{3n}\ge n^nn^nn^n\ge n^nn!n!=n^n(n!)^2$$