Question 527962
The induction basically claims that if for some k > 1, if


*[tex \LARGE 2^k \ge k^2], then


*[tex \LARGE 2^{k+1} \ge (k+1)^2].


Since *[tex \LARGE 2^{k+1} = 2(2^k) \ge 2k^2], we can construct the inequality


*[tex \LARGE 2^{k+1} \ge 2k^2 \ge (k+1)^2]


The inequality *[tex \LARGE 2k^2 \ge (k+1)^2] is equivalent to *[tex \LARGE 2k^2 \ge k^2 + 2k + 1 \Rightarrow k^2 - 2k - 1 \ge 0] It factors to *[tex \LARGE (k-1)^2 - 2 \ge 0 \Rightarrow (k-1)^2 \ge 2], which is true for sufficiently large k (it doesn't hold for k = 0,1,2 but we can check these cases separately).