How can I prove that 6n^2+10 ≥ 3(n+1)^2 given that n≥8?

Subtract the right side

.. 6n^2 + 10 - 3(n+1)^2 ≥ 0

.. = 3n^2 - 6n + 7 ≥ 0

.. = 3(n-1)^2 + 4 ≥ 0 … rewrite in vertex form, always true.

This expression is always positive, so the left side of the original inequality is always greater than the right side (for any n, not just n≥8).

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In the graphic, the blue curve is the left side of the inequality; the yellow curve is the right side.

6n^2 + 10

3(n+1)^2 = 3n^2 + 6n + 3

So, 3n^2 - 6n + 7 > 0, if and only if n > 8.

Can you show this equivalent statement

Solve the inequation, and show that the solution interval contains the interval [8, +oo[.