Prove that $1+ \frac{1}{x^4} \geq \frac{1}{x} + \frac{1}{x^3}$

Question

Prove That
$$1+ \frac{1}{x^4} \geq \frac{1}{x} + \frac{1}{x^3}$$

where $x \in \mathbb Z^{+}$

Answer 1

If $x \geq 2$, then $\frac{1}{x} + \frac{1}{x^3} \leq \frac{1}{2} + \frac{1}{2} = 1 < 1 + \frac{1}{x^4}$.

Answer 2

If $x \in \mathbb{Z}^+$, then $1 + \frac{1}{x^4} \geq \frac{1}{x} + \frac{1}{x^3} \iff x^4 + 1 \geq x^3 +x \iff x^3(x-1) \geq x-1$

Answer 3

Multiply by $x^4$ to get$$x^4+1\ge x^3+x \\ x^4-x^3\ge x-1 \\ x^3(x-1)\ge x-1\\ \ \ \ x^3\ge1.$$

Answer 4

Observe that the desired inequality is equivalent to $$1-\frac1{x^3}\ge\frac1x-\frac1{x^4},$$ which can be rewritten $$1-\frac1{x^3}\ge \frac1x\cdot\left(1-\frac1{x^3}\right).$$ Hence, we can rewrite it equivalently as $$\left(1-\frac1x\right)\left(1-\frac1{x^3}\right)\ge 0.\tag{$\star$}$$ Can you prove $(\star)$ for all nonzero real $x$? (It doesn't actually matter whether $x$ is an integer, or even positive, just nonzero real!) Or, more simply, that $(\star)$ holds for all positive real $x$? Or, even more simply, for all $x\ge1$? About the only thing that the assumption $x\in\Bbb Z^+$ does for us is allow an induction proof, which is unnecessary.

Answer 5

If $x=1$, then $1+\dfrac{1}{x^4}=\dfrac1x+\dfrac{1}{x^3}=2$. Otherwise, $x\ge2$, in which case $\dfrac{1}{x}\le\dfrac12$ and $\dfrac{1}{x^3}\le\dfrac18$, so $\dfrac{1}{x}+\dfrac{1}{x^3}\le\dfrac58<1<1+\dfrac{1}{x^4}$

Answer 6

Since $x$ is positive, this is equivalent to showing that $$ x^4+1\ge x^3+x $$ or $$ x^4-x^3-x+1\ge0 $$ Since $$ x^4-x^3-x+1=x^3(x-1)-(x-1)=(x^3-1)(x-1)=(x-1)^2(x^2+x+1) $$ you can probably go on.

Answer 7

My first instinct was to multiply by $x^2$ ($\geq 0$) to get a nice symmetric form $$ x^2+ \frac{1}{x^2} \geq x + \frac{1}{x}. $$

This is true, because $f(x) = x + \frac{1}{x}$ is nondecreasing ($f'(x) = 1 - \frac{1}{x^2}\geq 0$), and $x^2 \geq x$ for $x\in\mathbb{Z}_+$. Apparently I was lucky and this nice form was also easy to work with.