This particular question looks like Bernoulli's inequality. I've looked at similar questions but am still stuck on this problem: Show that $$ (1+x)^n > \frac {n(n-1)} 2 x^2 $$ for $x > 0$ and all positive integers $n$
The base case is obvious enough, then using the assumption
$$ (1+x)^k > \frac {k(k-1)} 2 x^2 $$
I also know I'm using this to prove $$(1+x)^{k+1} > \frac {(k+1)k} 2 x^2 = \frac {k(k-1)} 2 x^2 + kx^2$$
I'm multiplying by $(1+x)$ to use the inductive step
$$ (1+x)^{k+1} > \frac {k(k-1)} 2 x^2 + \frac {k(k-1)} 2 x^3 $$ which is not clear enough to me.
Continuing your proof starting from the inductive step:
We need to prove that $(1+x)^k(1+x) = (1+x)^k + x(1+x)^k > \dfrac{k(k-1)}{2}x^2 + kx^2$
Knowing that $(1+x)^k > \dfrac{k(k-1)}{2}x^2$ we are left to show that $x(1+x)^k > kx^2$
$x(1+x)^k = x\sum_0^k{{k}\choose{i}}x^i = \sum_0^k{{k}\choose{i}}x^{i+1} = x + kx^2 + \sum_3^k{{k}\choose{i}}x^{i+1} > kx^2$