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Let $x \geq 1$. Then is it true that $2x^3 - 3x^2 + 2 \geq 1$?

If so, how can I show this using only elementary ideas such as factorisation?

Of course, I can demonstrate this using the methods of single variable calculus. The outline is as follows: Let $f(x) \colon= 2x^3 - 3x^2 + 2$ for $x \geq 1$. Then $f^{\prime}(x) = 6x^2 - 6x = 6x(x-1) > 0$ for $x > 1$; so $f$ is an increasing function, and $f(1) = 1$.

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3 Answers 3

up vote 1 down vote accepted

We have $$ 2x^3-3x^2+1 = (x-1)(2x^2-x-1)$$ and $x-1\ge 0$ and $2x^2-x-1\ge 2x-x-1=x-1\ge 0$.

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Hint

Asking $$2x^3 - 3x^2 + 2 \geq 1$$ is identical to $$2x^3 - 3x^2 + 1 \geq 0$$ The cubic polynomial exhibits an obvious solution for $x=1$. Perform the polynomial division to get a quadratic, solve for the roots and decide where the lhs is positive.

I am sure that you can take from here.

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Yet another way: $$2x^3 - 3x^2 + 2 \ge 1 \iff x^3+x^3+1 \ge 3x^2$$

which is true $\forall x \ge 0$ by AM-GM, with equality only when $x=1$.

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