Prove for every positive real number $x$ that $1+\frac{1}{x^4}\geq\frac{1}{x}+\frac{1}{x^3}$

Question

Prove for every positive real number $x$ that $1+\frac{1}{x^4}\geq\frac{1}{x}+\frac{1}{x^3}$.

So for my sketched out proof I have: $1+\frac{1}{x^4}-\frac{1}{x}-\frac{1}{x^3}\geq 0$. We need to find a common denominator so I used $x^4$. I then got $\frac{x^4+1-x^3-x}{x^4}\geq 0$. I'm having trouble with this because I wanted to clear out the denominator but I can't find a way to do this.

Answer 1

Now, $x^4>0$ and $$x^4-x^3-x+1=x^3(x-1)-(x-1)=(x-1)^2(x^2+x+1)=$$ $$=(x-1)^2\left(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right)\geq0.$$

Answer 2

Here is another way.

$$1 + \frac{1}{x^4} \geq \frac{1}{x} + \frac{1}{x^3} \Longleftrightarrow 1 - \frac{1}{x} \geq \frac{1}{x^3}(1 - \frac{1}{x}).$$

When $x > 1$, $1 / x^3 < 1$. When $x < 1$, $1 / x^3 > 1$.

Answer 3

$$x^4+1-x^3-x=x^3(x-1)-(x-1)=(x-1)(x^3-1)=(x-1)^2(x^2+x+1)$$

Now $x^2+x+1=\dfrac{(2x+1)^2+3}4\ge\dfrac34$ for real $x$

Answer 4

Prove: $1+{1\over{x^4}}\ge{1\over{x}}+{1\over{x^3}}, x\in{\mathbb{R^+}}$

$x^4+1\ge{{x^3}+x}\Longleftrightarrow x^4+1-x^3-x\ge 0$.$\space$ The case for $x=1$ is trivial.

$x\gt 1:x^4-x^3-x+1=x(x^3-x^2-1)+1\gt x^3-x^2-1+1=x^3-x^2\gt 0$

$0\lt x\lt 1:x={1\over{a}},\space 1\lt a\in{\mathbb{Z^+}}$ in $1+{1\over{x^4}}-{1\over{x}}-{1\over{x^3}}\ge 0$ and proceed as above.