IMC 2006 · Day 1 · P5

Without loss of generality $a \ge b \ge c$, $d \ge e$. Let $c^2 = e^2 + \Delta$, $\Delta \in \mathbb{R}$. Then $d^2 = a^2 + b^2 + \Delta$ and the second equation implies $$a^4 + b^4 + (e^2 + \Delta)^2 = (a^2 + b^2 + \Delta)^2 + e^4, \qquad \Delta = -\frac{a^2 b^2}{a^2 + b^2 - e^2}.$$ (Here $a^2 + b^2 - e^2 \ge \frac{2}{3}(a^2 + b^2 + c^2) - \frac{1}{2}(d^2 + e^2) = \frac{1}{6}(d^2 + e^2) > 0$.)

Since $c^2 = e^2 - \frac{a^2 b^2}{a^2 + b^2 - e^2} = \frac{(a^2 - e^2)(e^2 - b^2)}{a^2 + b^2 - e^2} > 0$ then $a > e > b$.

Therefore $d^2 = a^2 + b^2 - \frac{a^2 b^2}{a^2 + b^2 - e^2} < a^2$ and $a > d \ge e > b \ge c$.

Consider a function $f(x) = a^x + b^x + c^x - d^x - e^x$, $x \in \mathbb{R}$. We shall prove that $f(x)$ has only two zeroes $x = 2$ and $x = 4$ and changes the sign at these points. Suppose the contrary. Then Rolle's theorem implies that $f'(x)$ has at least two distinct zeroes. Without loss of generality $a = 1$. Then $f'(x) = \ln b \cdot b^x + \ln c \cdot c^x - \ln d \cdot d^x - \ln e \cdot e^x$, $x \in \mathbb{R}$. If $f'(x_1) = f'(x_2) = 0$, $x_1 < x_2$, then $$\ln b \cdot b^{x_i} + \ln c \cdot c^{x_i} = \ln d \cdot d^{x_i} + \ln e \cdot e^{x_i}, \quad i = 1, 2,$$ but since $1 > d \ge e > b \ge c$ we have $$\frac{(-\ln b) \cdot b^{x_2} + (-\ln c) \cdot c^{x_2}} {(-\ln b) \cdot b^{x_1} + (-\ln c) \cdot c^{x_1}} \le b^{x_2 - x_1} < e^{x_2 - x_1} \le \frac{(-\ln d) \cdot d^{x_2} + (-\ln e) \cdot e^{x_2}} {(-\ln d) \cdot d^{x_1} + (-\ln e) \cdot e^{x_1}},$$ a contradiction. Therefore $f(x)$ has a constant sign at each of the intervals $(-\infty, 2)$, $(2, 4)$ and $(4, \infty)$. Since $f(0) = 1$ then $f(x) > 0$, $x \in (-\infty, 2) \cup (4, \infty)$ and $f(x) < 0$, $x \in (2, 4)$. In particular, $f(3) = a^3 + b^3 + c^3 - d^3 - e^3 < 0$.