I need to prove $n^4 - 5n \geq 2n^2 -6$, for all $n≥1$ by mathematical induction.

Question

I need to prove $n^4 - 5n \geq 2n^2 -6$, for all $n≥1$ by mathematical induction.

Base case holds.

Assumption

Assume holds for some fixed value in N called k, i.e.:

$k^4 - 5k \geq 2k^2 -6$

Induction step

Add 4k+2 both sides, induction assumption

$k^4 - 5k + (4k+2) \geq 2k^2 -6 + (4k+2)$

Rearrange $$k^4 - 5k + (4k+2) \geq (2k^2+4k+2) -6$$ $$k^4 - 5k + (4k+2) \geq 2(k^2+2k+1) -6$$ $$k^4 - 5k + (4k+2) \geq 2(k+1)^2 -6$$ $$k^4 - k+2 \geq 2(k+1)^2 -6$$

$$k^4 - (k -2) \geq 2(k+1)^2 -6$$

But $$(k+1)^4 - 5(k+1) \gt k^4 - (k -2)$$ as $ (k+1)^4 \gt k^4, 5k+5\gt k-2$

Therefore $$(k+1)^4 - 5(k+1) \gt 2(k+1)^2 -6 $$ $$(k+1)^4 - 5(k+1) \geq 2(k+1)^2 -6 $$

since the latter is a weaker statement.

Therefore holds for k+1.

What is wrong with this proof?

Answer 1

$n^4-5n\ge 2n^2-6$

$(n+1)^4-5(n+1)\ge 2(n+1)^2-6 ?$

$n^4+4n^3+6n^2+4n+1 -5n-5 \ge 2n^2+4n-4 ?$

$n^4-5n+ (4n^3+6n^2 -4)\ge 2n^2-6 + (2)$

$4n^3+6n^2-4\ge 2$

$2n^3+3n^2\ge 3$

$n^2(2n+3)\ge 3$

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$n>1\implies n^2\ge1 \land (2n+3)\ge 3$

$n^2(2n+3)=2n^3+3n^2\ge 3$

$4n^3+6n^2\ge 6$

$4n^3+6n^2-4\ge 2$

$4n^3+6n^2+4n-4\ge 4n+2$

Base case (n=1):$\ n^4-5n\ge 2n^2-6$

$n^4 +4n^3+6n^2-n-4 \ge 2n^2+4n-4$

$(n+1)^4-5n-5 \ge 2n^2+4n-4$

$(n+1)^4-5(n+1) \ge 2(n+1)^2-6$

It's true for $n=1$ and if it is true for $n$ it is true for $n+1$ so the conclusion follows by induction.

Answer 2

Suppose $n^4-5n\ge 2n^2-6$ or $n^4- 2n^2-5n+6 \ge 0$.

Then

$\begin{array}\\ (n+1)^4- 2(n+1)^2-5(n+1)+6 &=n^4+4n^3+6n^2+4n+1 -2n^2-4n-2-5n-5+6\\ &=n^4-2n^2-5n+6+4n^3+6n^2+4n+1 -4n-7\\ &\ge 4n^3+6n^2+4n+1 -4n-7 \qquad\text{(Induction hypothesis)}\\ &\ge 4n^3+6n^2-6\\ &\gt 0 \qquad \text{(For } n \ge 1)\\ \end{array} $