## Monday, February 13, 2006

### Function Substitution. Topic: Inequalities. Level: Olympiad.

Theorem: (Jensen's Inequality) Let $f$ be a function that is convex on the interval $I$. Given positive reals $a_1, a_2, \ldots, a_n \in I$ and positive real weights $w_1, w_2, \ldots, w_n$ such that $w_1+w_2+\cdots+w_n = 1$, we have

$w_1f(a_1)+w_2f(a_2)+\cdots+w_nf(a_n) \ge f(w_1a_1+w_2a_2+\cdots+w_na_n)$.

If $f$ is concave on $I$, the inequality is reversed.

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Problem: (1998 IMO Shortlist - A3) Let $r_{1},r_{2},\ldots ,r_{n}$ be real numbers greater than or equal to $1$. Prove that

$\frac{1}{r_{1} + 1} + \frac{1}{r_{2} + 1} + \cdots +\frac{1}{r_{n}+1} \geq \frac{n}{ \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}$.

Solution: The $n$ functions on the LHS and the single function on the RHS gives us the idea of using Jensen's. However, we can't really get a geometric mean via Jensen's, so this approach doesn't look promising.

On the other hand, since we have $r_i \ge 1$, consider the substitution $r_i = e^{a_i}$ for nonnegative reals $a_1, a_2, \ldots, a_n$. We then have $\sqrt[n]{r_1r_2 \cdots r_n} = e^{\frac{a_1+a_2+\cdots+a_n}{n}}$. Aha! We have converted the geometric mean to an arithmetic one, suggesting Jensen's.

Consider the function $f(x) = \frac{1}{e^x+1}$. Since $f^{\prime\prime}(x) = \frac{2xe^x}{(e^x+1)^2} \ge 0$ for $x \in [0, \infty)$, the function is convex and we may apply Jensen's on $f$. Setting $w_1 = w_2 = \cdots = w_n = \frac{1}{n}$, we have

$\frac{1}{n}f(a_1)+\frac{1}{n}f(a_2)+\cdots+\frac{1}{n}f(a_n) \ge f\left(\frac{a_1+a_2+\cdots+a_n}{n}\right)$,

which is just

$\frac{1}{r_1+1}+\frac{1}{r_2+1}+\cdots+\frac{1}{r_n+1} \ge \frac{n}{ \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}$,

as desired. QED.

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