# Problem 4186 from Crux Mathematicorum

### Solution 1

If $\displaystyle \int_0^1f(x)dx=0,$ then, since $f$ is continuous and $f(x)\ge 0,$ for $x\in [0,1],$ $f\equiv 0$ and hence the inequality stands. The same conclusion holds if $\displaystyle \int_0^1g(x)dx=0.$ So we assume that

$\displaystyle I=\int_0^1f(x)dx\gt 0$ and $\displaystyle J=\int_0^1g(x)dx\gt 0.$

The required inequality is equivalent to

$\displaystyle \int_0^1h(x)\left(\frac{f(x)}{I}-\frac{g(c)}{J}\right)dx\ge 0.$

Define the function $\phi:\,[0,1]\rightarrow\mathbb{R}$ as $\displaystyle \phi(x)=\frac{f(x)}{I}-\frac{g(c)}{J},$ $x\in [0,1].$ Clearly, $\phi$ is continuous and convex. Moreover,

$\displaystyle \int_0^1\phi(x)dx=\frac{1}{I}\int_0^1f(x)dx-\frac{1}{J}\int_0^1g(x)dx=1-1=0,$

and also $\phi(0)=0.$ Now, the required inequality is equivalent to $\displaystyle \int_0^1h(x)\phi(x)dx\ge 0.$

If $\phi$ is increasing on $[0,1],$ then, since $\phi(0)=0$ and $\displaystyle \int_0^1\phi(x)dx=0,$ we deduce $\phi\equiv 0$ and we are done. A similar conclusion holds if $\phi$ is decreasing. Otherwise, since $\phi$ is convex, we know that there is $t\in (0,1)$ such that $\phi$ is decreasing on $[0,t]$ and increasing on $[t,1].$ Thus it follows that $\phi(1)\ge 0$ and that there is $c\in (),1)$ such that $\phi(x)\le 0$ for $x\in [0,c]$ and $\phi(x)\ge 0$ for $x\in [c,1].$ Then, for $x\in [0,c],$ $h(x)-h(c)\le 0$ and, therefore, $h(x)\phi(x)\ge h(c)\phi(x),$ for $x\in [0,c].$ Similarly, $h(x)\phi(x)\ge h(c)\phi(x),$ for $x \in [c,1].$ Thus, $h(x)\phi(x)\ge h(c)\phi(x)$ on $[0,1]$ and

$\displaystyle \int_0^1h(x)\phi(x)dx\ge\int_0^1h(c)\phi(x)dx=0,$

which completes the proof.

### Solution 2

Integrating by parts and capitalizing the cumulative function:

$lhs=\int_0^1 g(x) h(x) \, dx\, \times \int_0^1 f(x) \, dx =F(1) \left(G(1) h(1)-\int_0^1 g(x) h'(x) \, dx\right)$

$rhs=\int_0^1 f(x) h(x) \, dx \,\times \int_0^1 g(x) \, dx=G(1) \left(F(1) h(1)-\int_0^1 f(x) h'(x) \, dx\right)$

$rhs-lhs=F(1) \int_0^1 g(x) h'(x) \, dx-G(1) \int_0^1 f(x) h'(x) \, dx \; \text{ with }h'(x)\geq 0$

We multiply both sides with a constant $\lambda$ such as $\lambda \int_0^1 h'(x) \, dx=1$.

Owing to concavity

$\lambda \int_0^1 g(x) h'(x) \, dx \leq G(1)$

and convexity

$\lambda \int_0^1 f(x) h'(x) \, dx \geq F(1)$

QED

### Acknowledgment

This is Problem 4186 from the Canadian Crux Mathematicorum; the problem was posed by Florin Stanescu. Leo Giugiuc was the only one, besides the proposer, who submitted a correct solution. His solution was featured in the magazine. I am grateful to Leo Giugiuc who kindly shared with me the problem and his solution (Solution 1). Solution 2 is by N. N. Taleb.