I saw, flipping through a list of problems for the Mathematics Olympiad for Juniors, the following question:
"If $0<x<1$ prove that $x-x^3\;<\frac{1}{2}$."
I would be curious what the simplest solution would be. I had another post like this here. And I mean a solution that does not call for advanced mathematical knowledge, for example Calculus.
Anyone with even a little Calculus experience realizes that the statement of the problem hides an insufficiency: does the expression $x-x^3$ have a maximum or a minimum on the interval $0<x<1$ ?; and what are the values of $x$ that achieve them ?
Following this line, Calculus teaches us to find out the largest and smallest value taken by the function
$$f\;:\;[0;1]\rightarrow \mathbb{R},\;\;f(x)=x-x^3$$
(or the extension of this function to the entire real axis $-\infty <x<+\infty$).
The derivative ${f}'=1-3x^2$ has critical (stationary) points $x_\pm=\pm \frac{1}{\sqrt{3}}$.
As can be seen from the Variation Table of the Function, at $x_-$ we have a local minimum, and at $x_+$ - a local maximum. The conclusion of this study is that
If $0<x<1$ then $x-x^3 \leqslant \frac{2}{3\sqrt{3}}. \;\;\;\;\;(1)$
Equality occur at $x_+=\frac{1}{\sqrt{3}}$.
Remark CiP Before I had time to complete the Variation Table of the function, Mr. George Stoica already posted his comment with an elementary solution to both the initial version (obtaining a slight strengthening) and a solution to the "optimal" version (1).
The title of the post is inspired by the similar title of a book, we could call it famous, by Ivan NIVEN, (The MATHEMATICAL ASSOCIATION of AMERICA, 1981). Even after decades, this book is still quoted with interest; see Berkeley Math Circle : Tom Rike, Maxima and Minima Problems WITHOUT Calculus, 2002
We will give a demonstration based on the maximum product principle. It is stated as follows
If the sum of $n$ non-negative real numbers is constant, then their
product is maximum when the numbers are equal.
Let be the product of three factors $P=x\cdot (1-x) \cdot (1+x).$
The sum of the three factors is $x+(1-x)+(1+x)=2+x$. Unfortunately, it is not constant, so we will resort to a "trick". We consider the product
$$P_1=(\sqrt{3}+1)(2+\sqrt{3}) \cdot P=$$
$$=[(\sqrt{3}+1)x]\cdot [(2+\sqrt{3})(1-x)] \cdot [1+x].$$
The sum of the three factors is now
$[x\sqrt{3}+x]+[2+\sqrt{3}-2x-x\sqrt{3}]+[1+x]=3+\sqrt{3}$
and is now a constant. So the maximum of $P_1$ occurs when
$$(\sqrt{3}+1)x=(2+\sqrt{3})(1-x)=1+x.$$
But the three conditions above are satisfied exactly when $x=\frac{1}{\sqrt{3}}.$
Now, $P_1$ and $P$ differ by a positive factor, so they are maximum at the same time.
$\blacksquare$
Things don't end there. Some might say that this solution is like pulling a rabbit out of a hat. I applied a method known in the French mathematical literature as belonging to M. Grillet. Unfortunately, in the "Éléments d'algèbre, à l'usage des candidats au baccalauréat ès sciences et aux écoles spéciales" par Eugène Rouché (Paris, MALLET-BACHELIER , IMPRIMEUR-LIBRAIRE, 1857) on page 181, only the Nouvelles Annales are mentioned, without any other specification.
We start by finding three numbers $\alpha,\;\beta,\;\gamma\;$ so that the product
$$P_1=\alpha \cdot \beta \cdot \gamma \cdot P=[\alpha \cdot x]\cdot [\beta \dot (1-x)] \cdot [\gamma \cdot (1+x)]$$
has a constant sum of factors. The sum of these factors is
$$(\alpha x)+(\beta-\beta x)+(\gamma +\gamma x)=(\alpha -\beta+\gamma)\cdot x+(\beta+\gamma).$$
If we choose $\alpha -\beta +\gamma =0$, then $\alpha = \beta-\gamma$ and
$$P_1=[(\beta-\gamma)x]\cdot [\beta(1-x)]\cdot [\gamma(1+x)]$$
has a constant sum of factors. $P_1$ is maximum for
$$(\beta-\gamma) \cdot x =\beta-\beta\cdot x=\gamma+\gamma \cdot x .\tag{2}$$
From the first equality we get $x=\frac{\beta}{2\beta-\gamma}$ and from the second equality we get $x=\frac{\beta-\gamma}{\beta+\gamma}.$ In order for relations (2) to be compatible, we must have
$$\frac{\beta-\gamma}{\beta+\gamma}=\frac{\beta}{2\beta-\gamma}\;\Leftrightarrow\;\beta^2-4\beta \gamma+\gamma^2=0.$$
We choose $\frac{\beta}{\gamma}>1$ to have $\beta-\gamma >0$, so $\frac{\beta}{\gamma}=2+\sqrt{3}.$ In our trick we took $\gamma=1,\;\beta=2+\sqrt{3},\;\alpha=1+\sqrt{3}.$
(Remark. In the case of more than three variables, conditions like those in (2) lead to equations of degree higher than 2, so they are more difficult to handle.)
!! Finally, I found exactly. Grillet, H. Méthode élémentaire pour résoudre quelques questions sur les maximums. Nouvelles annales de mathématiques : journal des candidats aux écoles polytechnique et normale, Serie 1, Volume 9 (1850), pp. 70-73.
