0
$\begingroup$

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?

$\endgroup$
  • 6
    $\begingroup$ This isn't true. At $x=1$ you have $2 \geq 4$. $\endgroup$ – Sloan Sep 19 '16 at 20:09
  • 1
    $\begingroup$ This would be a better problem is it were $x^4+3$. First, it would be true and second, equality is achieved at $x=1$. $\endgroup$ – B. Goddard Sep 19 '16 at 20:50
4
$\begingroup$

$f(x)=x^4+1$ 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 $x=1$.

$\endgroup$
1
$\begingroup$

Set $x = 1$

$$2 \geq 4$$

Your statement is not valid for all the reals.

$\endgroup$
  • $\begingroup$ Ah sorry should change my tag.. $\endgroup$ – bigfocalchord Sep 19 '16 at 20:11
0
$\begingroup$

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 $$c=\frac 13\sqrt[3]{216-24\sqrt{78}}$$ $$d=\frac{2\sqrt[3]{9+\sqrt{78}}}{\sqrt[3]{9}}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.