**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.

Without the weights on the RHS you're really applying Karamata's, did you mean to leave them out? XD

ReplyDeleteAlso, 1/(x+1) is decreasing, but it's convex.

Oops. Wrote it in the wrong form.

ReplyDeleteMeh, you're right, 1/(x+1) is convex, but it doesn't work since you get the arithmetic instead of geometric mean.