1974 IMO Problems/Problem 5

Problem 5

Determine all possible values of $S = \frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{a+c+d}$ where $a, b, c, d,$ are arbitrary positive numbers.

Solution

Note that $2 = \frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{c+d}+\frac{d}{c+d} > S > \frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d} = 1.$ We will now prove that $S$ can reach any range in between $1$ and $2$ .

Choose any positive number $a$ . For some variables such that $k, m, l > 0$ and $k + m + l = 1$ , let $b = ak$ , $c = am$ , and $d = al$ . Plugging this back into the original fraction, we get $S = \frac{a}{a+ak+al}+\frac{ak}{a+ak+am}+\frac{am}{ak+am+al}+\frac{al}{a+am+al} = \frac{1}{1+k+l}+\frac{k}{1+k+m}+\frac{m}{k+m+l}+\frac{l}{1+m+l}.$ The above equation can be further simplified to $S = \frac{1}{2-m}+\frac{k}{2-l}+m+\frac{l}{2-k}.$ Note that $S$ is a continuous function and that $f(m) = m + \frac{1}{2-m}$ is a strictly increasing function. We can now decrease $k$ and $l$ to make $m$ tend arbitrarily close to $1$ . We see $\lim_{m\to1} m + \frac{1}{2-m} = 2$ , meaning $S$ can be brought arbitrarily close to $2$ . Now, set $a = d = x$ and $b = c = y$ for some positive real numbers $x, y$ . Then $S = \frac{2x}{2x+y} + \frac{2y}{2y+x} = \frac{2y^2 + 8xy + 2x^2}{2y^2 + 5xy + 2x^2}.$ Notice that if we treat the numerator and denominator each as a quadratic in $y$ , we will get $1 + \frac{g(x)}{2y^2 + 5xy + 2x^2}$ , where $g(x)$ has a degree lower than $2$ . This means taking $\lim_{y\to\infty} 1 + \frac{g(x)}{2y^2 + 5xy + 2x^2} = 1$ , which means $S$ can be brought arbitrarily close to $1$ . Therefore, we are done. $\[\]$ ~Imajinary

1974 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
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1974 IMO Problems/Problem 5

Problem 5

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