Proof by induction that $\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$ I am a bit new to logical induction, so I apologize if this question is a bit basic.

I tried proving this by induction:

$$\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$$

Starting with the base case $n=1$:

$$1^2\ge1^2$$

Then to prove that $P(n)\to P(n+1)$:

$$\left(\sum_{k=1}^{n+1}k\right)^2\ge\sum_{k=1}^{n+1}k^2$$
$$\left((n+1)\sum_{k=1}^nk\right)^2\ge(n+1)^2\sum_{k=1}^nk^2$$
$$(n+1)^2\left(\sum_{k=1}^nk\right)^2\ge(n+1)^2\sum_{k=1}^nk^2$$

Since they're both multiplied by $(n+1)^2$, it's easy to see that if $\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$, then $(n+1)^2\left(\sum_{k=1}^nk\right)^2\ge(n+1)^2\sum_{k=1}^nk^2$. But if the $\ge$ were replaced with a $\le$, the proof would still be valid, even though it's demonstrably not true. 

I can see that it would take strong induction would fix this problem, but if I didn't know by observation that $\left(\sum_{k=1}^nk\right)^2\le\sum_{k=1}^nk^2$ isn't true by observation, how would I know to use strong induction?
Hints:

$$\sum_{k=1}^n k=\frac{n(n+1)}2$$

$$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}6$$
$$\sum^n_1 k=1+2+ \dots n$$

$$(\sum^n_1 k)^2=(1+2+ \dots n)^2 \neq1^2+2^2 + \dots n^2$$