## Monday, March 27, 2006

### Yum! Topic: Inequalities. Level: Olympiad.

Problem: Let $a,b,c$ be positive reals such that $a^2+b^2+c^2 = 3$. Prove that

$\frac{a}{a^3+a^2+4}+\frac{b}{b^3+b^2+4}+\frac{c}{c^3+c^2+4} \le \frac{1}{2}$.

Solution: We will use a method of solving inequalities called "Isolated Fudging." Since the variables are separated we will prove an inequality for each individual term. We guess that there exists a real $k$ such that

$\frac{a}{a^3+a^2+4} \le \frac{1}{6}+ka^2-k$.

Doing the same for each variable and summing up, we see that this will prove the result. To find $k$, we throw in some calculus (the part ensuing is not necessary in a formal solution write-up, so you don't need to include any calculus in the "official" solution).

Let $f(a) = \frac{1}{6}+ka^2-k-\frac{a}{a^3+a^2+4}$. Differentiating with respect to $a$, we have

$f^{\prime}(a) = 2ak-\frac{(a^3+a^2+4)-a(3a^2+2a)}{(a^3+a^2+4)^2}$.

Looking at the equality case in the problem, we see that $a = b = c = 1$. So if our guess is correct, then $f^{\prime}(1) = 0$. So

$f^{\prime}(1) = 2k-\frac{1}{36} = 0 \Rightarrow k = \frac{1}{72}$.

Now back to the solution. It remains to show that

$\frac{a}{a^3+a^2+4} \le \frac{1}{6}+\frac{1}{72}a^2-\frac{1}{72} = \frac{11}{72}+\frac{1}{72}a^2$,

which is equivalent to

$72a \le (a^3+a^2+4)(a^2+11)$,

or

$0 \le (a-1)^2(a^3+3a^2+16a+44)$.

Since $a$ is a positive real, we have $a^3+3a^2+16a+44 > 0$ and $(a-1)^2 \ge 0$, so it is true. Hence

$\frac{a}{a^3+a^2+4}+\frac{b}{b^3+b^2+4}+\frac{c}{c^3+c^2+4} \le \left(\frac{11}{72}+\frac{1}{72}a^2\right) + \left(\frac{11}{72}+\frac{1}{72}b^2\right) + \left(\frac{11}{72}+\frac{1}{72}c^2\right) = \frac{1}{2}$,

as desired. QED.

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Comment: Isolated Fudging is a pretty neat trick and there are many variations on it, which are nice to learn. It lets you come up with solutions that have extremely strange numbers but magically work out.

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Practice Problem: (2005 IMO - #3) Prove that for all positive $a,b,c$ with product at least $1$,

$\frac{a^5-a^2}{a^5+b^2+c^2} + \frac{b^5-b^2}{b^5+c^2+a^2} + \frac{c^5-c^2}{c^5+a^2+b^2} \ge 0$.