# Prove that $x^4 + 1 ≥ 4x$ for all real numbers x.

Prove that $x^4 + 1 ≥ 4x$ for all real numbers x.

What I have attempted:

Consider $$x^4+1≥ 4x$$

$$\Leftrightarrow x^4+1+2x^2 \geq 4x+2x^2$$

$$\Leftrightarrow (x^2+1)^2 \geq 2(x^2+2x+1) - 2$$

$$\Leftrightarrow (x^2+1)^2 - 2(x+1)^2 +2 \geq 0$$

Now this is where I am stuck, am I on the correct track?

• This isn't true. At $x=1$ you have $2 \geq 4$. – Sloan Sep 19 '16 at 20:09
• This would be a better problem is it were $x^4+3$. First, it would be true and second, equality is achieved at $x=1$. – B. Goddard Sep 19 '16 at 20:50

$f(x)=x^4+1$ is a convex function on $\mathbb{R}$. Let we consider the equation of the tangent line at $x=1$: it is $g(x)=4(x-1)+2$. $f(x)\geq g(x)$ implies:

$$x^4+1 \geq 4x-2.$$ As already remarked, the original inequality actually does not hold: just consider what happens at $x=1$.

Set $x = 1$

$$2 \geq 4$$

Your statement is not valid for all the reals.

• Ah sorry should change my tag.. – bigfocalchord Sep 19 '16 at 20:11

Too long for a comment:

The inequality is true for all $$\displaystyle x\in \mathbb{R} \,\backslash\, \left(\frac{a-b}{2}, \frac{a+b}{2}\right)$$ ($\frac{a-b}{2}\approx 0.251,\frac{a+b}{2}\approx 1.49$)

where $$a=\sqrt{c+d}$$ $$b=\sqrt{\frac{8}{\sqrt{c+d}}-(c+d)}$$ again where $$c=\frac 13\sqrt[3]{216-24\sqrt{78}}$$ $$d=\frac{2\sqrt[3]{9+\sqrt{78}}}{\sqrt[3]{9}}$$