Competition problem — Algebra — Maths Olympiad Prep

Algebra Difficulty 5.0 AIME Prove it Brazil

Determine the smallest real number $C$ such that the inequality
$C(x_1^{2005} + x_2^{2005} + x_3^{2005} + x_4^{2005} + x_5^{2005}) \geq x_1 x_2 x_3 x_4 x_5 (x_1^{125} + x_2^{125} + x_3^{125} + x_4^{125} + x_5^{125})^{16}$
holds for all positive real numbers $x_1, x_2, x_3, x_4, x_5$ .

Solution

We have
$5 (x_1^{2005} + x_2^{2005} + x_3^{2005} + x_4^{2005} + x_5^{2005}) \geq (x_1^5 + x_2^5 + x_3^5 + x_4^5 + x_5^5) (x_1^{2000} + x_2^{2000} + x_3^{2000} + x_4^{2000} + x_5^{2000})$
by Chebyshev. Also,
$x_1^5 + x_2^5 + x_3^5 + x_4^5 + x_5^5 \geq 5x_1x_2x_3x_4x_5$
by AM-GM and
$\frac{x_1^{2000} + x_2^{2000} + x_3^{2000} + x_4^{2000} + x_5^{2000}}{5} \geq \left( \frac{x_1^{125} + x_2^{125} + x_3^{125} + x_4^{125} + x_5^{125}}{5} \right)^{16}$
Combining these inequalities gives $C \leq 5^{15}$ . But substituting $x_1 = x_2 = x_3 = x_4 = x_5 = 1$ gives $C \geq 5^{15}$ . Thus $C = 5^{15}$ .

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