Mathematical Reflections 2017, Issue 2 - Problem S407

Problem:
Let $f(x)=x^3+x^2-1$. Prove that for any positive real numbers $a, b, c, d$ satisfying $$a+b+c+d>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d},$$ at least one of the numbers $af(b)$, $bf(c)$, $cf(d)$, $df(a)$ is different from $1$.

Proposed by Adrian Andreescu, Dallas, Texas, USA

Solution:
Assume by contradiction that $af(b)=bf(c)=cf(d)=df(a)=1$. The given inequality becomes
$$\begin{equation}\label{first-ineq} a+b+c+d>f(a)+f(b)+f(c)+f(d). \end{equation}$$
Observe that $f(x)$ is increasing on $(0,+\infty)$ and $f(x)<x$ if and only if $x<1$. Moreover, $f$ is convex on $(0,+\infty)$, so by inequality $\eqref{first-ineq}$ and Jensen's Inequality, we have $$\dfrac{a+b+c+d}{4}>\dfrac{f(a)+f(b)+f(c)+f(d)}{4} \geq f\left(\dfrac{a+b+c+d}{4}\right).$$ It follows that $\dfrac{a+b+c+d}{4}<1$, i.e. $a+b+c+d<4$ and from $\eqref{first-ineq}$, we get
$$f(a)+f(b)+f(c)+f(d)<4.$$ On the other hand, by the AM-GM Inequality, we have $$\begin{array}{lll} af(a)+bf(b)+cf(c)+df(d) & \geq & 4 \\ bf(a)+cf(b)+df(c)+af(d) & \geq & 4 \\ cf(a)+df(b)+af(c)+bf(d) & \geq & 4 \\ df(a)+af(b)+bf(c)+cf(d) & \geq & 4. \end{array}$$ Adding these four inequalities, we get $(a+b+c+d)(f(a)+f(b)+f(c)+f(d))\geq 16$, i.e. $$f(a)+f(b)+f(c)+f(d) \geq \dfrac{16}{a+b+c+d} > 4,$$ contradiction.
The conclusion follows.