How can I prove $2^n+10n^2 le 12 cdot 2^n, n ge 1$ is true by induction?

$2^n+10n^2 le 12 cdot 2^n, n ge 4^{**}$

Divide both sides by $2^n$ $$1+frac{10 n^2}{2^n}le 12;;nge 4tag{1}$$ For $n=4$ it is true $$1+frac{10cdot 4^2}{2^4}=11le 12$$ Now suppose $(1)$ is true for $n$ and let's prove it for $n+1$ $$1+frac{10 (n+1)^2}{2^{n+1}}=1+frac{1}{2}left(frac{10n^2}{2^n}right)+frac{1}{2}left(frac{20n+10^*}{2^n}right) <1+frac{1}{2}left(frac{10n^2}{2^n}right)+frac{1}{2}left(frac{10n^2}{2^n}right)=\=1+frac{10n^2}{2^n}le 12$$ $^*quad 20n+10<10n^2$ for $nge 4$

$^{**}$ inequality is false for $n=3$

indeed $$2^3+10cdot 3^2 le 12 cdot 2^3to 98le 96$$ which is false

We can use

$10n^2 - 20n - 10 = 10 [(n-1)^2 - 2] gt 0$ for $n gt 2$

i.e $20n^2 gt 10n^2 + 20n + 10 implies 10n^2 gt frac{1}{2} (10n^2 + 20n + 10)$

We have $2^n+10n^2 le 12 times 2^n$

and we need to show $2^{n+1}+10(n+1)^2 le 12 times 2^{n+1}$

$2^{n+1}+10(n+1)^2 = 2 [2^n + frac{1}{2}(10n^2 + 20n + 10)] lt 2(2^n + 10n^2)$

i.e $2^{n+1}+10(n+1)^2 leq 12 times 2^{n+1}$

$ k^2 > 2k+1 forall k>2$

$2k^2 > (k+1)^2$ $Rightarrow 10(k+1)^2 < 20k^2$ $2^{k+1} + 10(k + 1)^2 < 2^{k+1} + 20k^2$