<s>[INST] Prove $k^8 \geq (k^2 - 1)^4$ for $k \geq 1$ Can anyone please help me with following inequality:
$$k^8 \geq (k^2 - 1)^4, \text{where} \ k \geq 1.$$
I tried induction but get stuck at step $k=n+1$ and can't progress anywhere.
 [/INST] Since $k^8 = (k^2)^4$, we just need to show that (for $a,b\geq0$) $a\geq b \implies a^4 \geq b^4$. Indeed, if $a\geq b$, then $a-b\geq 0$, and so $a^4-b^4 = (a-b)(a^3+a^2b+ab^2+b^3) \geq 0$. Thus, $a^4 \geq b^4$.
Clearly, $k^2>k^2-1$, and so $k^8 > (k^2-1)^4$ since $k^2, k^2-1 \geq 0$.
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