Question

Find a positive number that exceeds its principal cube root by least amount?​

Answer 1

$\huge\underbrace\mathfrak\red{answer}$

Here is a method that does not require calculus:

Lemma: If x+y+z=kx+y+z=k (where kk is a constant), the product xyzxyz is maximized when x=y=z=k3x=y=z=k3

Proof:

By AM−GMAM−GM, we have

x+y+z3≥xyz−−−√3x+y+z3≥xyz3

k3≥xyz−−−√3k3≥xyz3

(k3)3≥xyz(k3)3≥xyz

We can see that the maximum of xyzxyz is (k3)3(k3)3, with equality occurring 

Answer 2

Answer:

Here is a method that does not require calculus:

Here is a method that does not require calculus:Lemma: If x+y+z=kx+y+z=k (where kk is a constant), the product xyzxyz is maximized when x=y=z=k3x=y=z=k3

Here is a method that does not require calculus:Lemma: If x+y+z=kx+y+z=k (where kk is a constant), the product xyzxyz is maximized when x=y=z=k3x=y=z=k3Proof:

Here is a method that does not require calculus:Lemma: If x+y+z=kx+y+z=k (where kk is a constant), the product xyzxyz is maximized when x=y=z=k3x=y=z=k3Proof:By AM−GMAM−GM, we have

Here is a method that does not require calculus:Lemma: If x+y+z=kx+y+z=k (where kk is a constant), the product xyzxyz is maximized when x=y=z=k3x=y=z=k3Proof:By AM−GMAM−GM, we havex+y+z3≥xyz−−−√3x+y+z3≥xyz3

Here is a method that does not require calculus:Lemma: If x+y+z=kx+y+z=k (where kk is a constant), the product xyzxyz is maximized when x=y=z=k3x=y=z=k3Proof:By AM−GMAM−GM, we havex+y+z3≥xyz−−−√3x+y+z3≥xyz3k3≥xyz−−−√3k3≥xyz3

Here is a method that does not require calculus:Lemma: If x+y+z=kx+y+z=k (where kk is a constant), the product xyzxyz is maximized when x=y=z=k3x=y=z=k3Proof:By AM−GMAM−GM, we havex+y+z3≥xyz−−−√3x+y+z3≥xyz3k3≥xyz−−−√3k3≥xyz3(k3)3≥xyz(k3)3≥xyz

Here is a method that does not require calculus:Lemma: If x+y+z=kx+y+z=k (where kk is a constant), the product xyzxyz is maximized when x=y=z=k3x=y=z=k3Proof:By AM−GMAM−GM, we havex+y+z3≥xyz−−−√3x+y+z3≥xyz3k3≥xyz−−−√3k3≥xyz3(k3)3≥xyz(k3)3≥xyzWe can see that the maximum of xyzxyz is (k3)3(k3)3, with equality occurring