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Algebra Difficulty 6.1 National olympiad Prove it

Example 5 Given that a,b,ca, b, c are positive real numbers, satisfying a+b+cabca+b+c \geqslant a b c. Prove: at least two of the following three inequalities hold
6a+3b+2c2,6b+3c+2a2,6c+3a+2b2\frac{6}{a}+\frac{3}{b}+\frac{2}{c} \geqslant 2, \frac{6}{b}+\frac{3}{c}+\frac{2}{a} \geqslant 2, \frac{6}{c}+\frac{3}{a}+\frac{2}{b} \geqslant 2 \text {. }

Solution

Prove by contradiction.
(i) If 6a+3b+2c17(1a+1b+1c)173>14\frac{6}{a}+\frac{3}{b}+\frac{2}{c}17\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 17 \sqrt{3}>14, contradiction!
Therefore, the conclusion holds.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.