as title, is there a way to find the minimum value of $x^4-x$ without using calculus?
By calculus it's easy as $(x^4-x)'=4x^3-1$, so we got $x=\frac1{\sqrt[3]4}$, then we get the min value. But without calculus, is it possible?
I tried to use AM-GM but to no avail…
Best Answer
Once you know the minimum must be negative [as $x^4-x=x(x^3-1) <0 $ when $x\in (0, 1)$], it isn't hard to use AM-GM. Let this minimum be $-m$.
$$x^4+m=x^4+3\times\frac13m \geqslant 4\sqrt[4]{x^4m^3/(3^3)}=4\sqrt[4]{\frac{m^3}{3^3}}|x| \geqslant 4\sqrt[4]{\frac{m^3}{27}}x$$ Hence if we select $m$ s.t. $m^3=\frac{27}{4^4}$, we get $x^4-x \geqslant -m$ and hence the minimum.
P.S. It is simple to see that equality is possible.