In mathematical competitions, many problems are specifically designed with a certain technique. To a large extent, these techniques are not particularly intuitive; in other words, it is difficult to follow them "logically," or the methods are not systematic. This is one reason why I somewhat dislike math competition problems. On the other hand, I personally believe that math competitions can train one’s thinking ability better than physics competitions, especially in terms of abstract thinking and geometric imagination. Therefore, working on some of these problems can still be beneficial.
Below is a classic competition problem, which appeared in the 42nd IMO held in South Korea:
Let a, b, c be positive real numbers. Prove that: \frac{a}{\sqrt{a^2+8bc}} + \frac{b}{\sqrt{b^2+8ac}} + \frac{c}{\sqrt{c^2+8ab}} \geq 1
I have seen several different proofs for this problem. On page 376 of International Mathematical Olympiad, two different proofs are provided. One of them is as follows:
Although the logic of the other proof is different, I still find it difficult to conceive. Yesterday, while browsing Shulian Tiandi, I discovered an exquisite proof that aligns well with conventional thinking. The main tool it utilizes is Jensen’s Inequality:
If a function f(x) satisfies f(a_1 x_1 + a_2 x_2) \geq a_1 f(x_1) + a_2 f(x_2) within an interval, it is called a concave function (or "upper convex" function), where a_1, a_2 are non-negative and a_1 + a_2 = 1. The concavity of a function can also be determined by its second derivative: if f''(x) < 0 within the interval, then the function is concave in that interval. (If the direction of the inequality is reversed, the function is called a convex function, or "lower convex" function).
Jensen’s Inequality actually generalizes this relationship: if a function satisfies f(a_1 x_1 + a_2 x_2) \geq a_1 f(x_1) + a_2 f(x_2) for any non-negative a_1 + a_2 = 1, then for any non-negative numbers a_1, a_2, \dots, a_n such that a_1 + a_2 + \dots + a_n = 1, we have: f(a_1 x_1 + a_2 x_2 + \dots + a_n x_n) \geq a_1 f(x_1) + a_2 f(x_2) + \dots + a_n f(x_n)
This is a powerful inequality that can be used to derive almost all other inequalities, such as the AM-GM inequality, the Cauchy-Schwarz inequality, and so on. Its proof is omitted here; the most direct method is using Taylor series expansion, and specific details can be found in online resources. Now, let’s return to the problem discussed in this article. The proof of this problem involves several commonly used methods for proving inequalities.
[The following section was struck out in the original post due to a flaw:]
First is form transformation, then mining hidden conditions.
Form transformation:\sout{\frac{1}{\sqrt{1+\frac{8bc}{a^2}}} + \frac{1}{\sqrt{1+\frac{8ac}{b^2}}} + \frac{1}{\sqrt{1+\frac{8ab}{c^2}}} \geq 1}
Mining hidden conditions:Let \frac{bc}{a^2}=x, \frac{ac}{b^2}=y, \frac{ab}{c^2}=z, then the expression becomes:\sout{\frac{1}{\sqrt{1+8x}} + \frac{1}{\sqrt{1+8y}} + \frac{1}{\sqrt{1+8z}} \geq 1}
where the hidden condition is xyz=1.
The next step is brilliant! Since using Jensen’s Inequality involves summation, while the given condition is in the form of a product, we want to turn it into a sum. Thus, we naturally think of exponential operations: let x=e^u, y=e^v, z=e^w, then the condition becomes u+v+w=0. The problem becomes:\sout{\frac{1}{\sqrt{1+8e^u}} + \frac{1}{\sqrt{1+8e^v}} + \frac{1}{\sqrt{1+8e^w}} \geq 1}
The most ingenious part of this transformation is that it not only performs a feasible substitution but also directly incorporates the inherent constraint of the problem: x, y, z are all positive!
Next, applying Jensen’s Inequality is very simple:\sout{\frac{1}{\sqrt{1+8e^u}} + \frac{1}{\sqrt{1+8e^v}} + \frac{1}{\sqrt{1+8e^w}} \geq \frac{3}{\sqrt{1+8e^{(\frac{u+v+w}{3})}}} = 1}(This is a concave function)
The proof originally provided in this article had a flaw: we cannot prove that \frac{1}{\sqrt{1+8e^x}} is concave over the entire set of real numbers \mathbb{R}. Therefore, the proof needs to be corrected:
We still use Jensen’s Inequality, but we apply it to the function f(x) = \frac{1}{\sqrt{x}}. It can be proven that f''(x) > 0 holds for all positive numbers x, so Jensen’s Inequality can be used (as a premise for a convex function).
It is not difficult to derive: \begin{aligned} &\left(\frac{a}{a+b+c}\right)\frac{1}{\sqrt{a^2+8bc}} + \left(\frac{b}{a+b+c}\right)\frac{1}{\sqrt{b^2+8ac}} + \left(\frac{c}{a+b+c}\right)\frac{1}{\sqrt{c^2+8ab}} \\ &\geq \frac{1}{\sqrt{\left(\frac{a}{a+b+c}\right)(a^2+8bc) + \left(\frac{b}{a+b+c}\right)(b^2+8ac) + \left(\frac{c}{a+b+c}\right)(c^2+8ab)}} \\ &= \sqrt{\frac{a+b+c}{a^3+b^3+c^3+24abc}} \end{aligned}
Thus: \frac{a}{\sqrt{a^2+8bc}} + \frac{b}{\sqrt{b^2+8ac}} + \frac{c}{\sqrt{c^2+8ab}} \geq \sqrt{\frac{(a+b+c)^3}{a^3+b^3+c^3+24abc}}
Now, we only need to prove (a+b+c)^3 \geq a^3+b^3+c^3+24abc. This can be easily derived using the AM-GM inequality, so I will not elaborate further.
The above proof is sourced from American Olympians.
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