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对于一类分式不等式的证明题,如果大胆将左、右两边“互相叠加”,兴许产生意料不到的奇迹! 定理1 欲证明不等式：P＞Q, 只须证明不等式：P+Q＞2Q。这个定理1太浅显了。例1 设a＞b＞c,求证：a~2/(a-b)+b~2/(b-c)＞a+2b+c。(第32届乌克兰数学竞赛试题) 证明设P=a~2/(a-b)+b~2/(b-c),Q=a+2b+c；考察新不等式：P+Q=(a~2/(a-b)+a-b)+(b~2/(b-c)+b-c)+(2b+2c)＞2a+2b+(2b+2c)=2(a+2b+c)=2Q,显然,P+Q＞2Q,依定理1,知P＞Q,故原不等式获证。 (注：此处不能取“=”,因为a~2/(a-b)+a-b≥2a,b~2/(b-c)+b-c≥2b等号不能同时成立)
For the proving question of a type of fractional inequality, if you boldly superimpose the left and right sides on each other, you may have an unexpected miracle! Theorem 1 To prove the inequality: P>Q, just prove the inequality: P+Q>2Q . This theorem 1 is too shallow. Example 1 Let a>b>c verify that a~2/(a-b)+b~2/(b-c)>a+2b+c. (Test of the 32nd Ukrainian Mathematical Contest) Proof Let P=a~2/(ab)+b~2/(bc), Q=a+2b+c; investigate the new inequality: P+Q=(a~2/ (ab)+ab)+(b~2/(bc)+bc)+(2b+2c)>2a+2b+(2b+2c)=2(a+2b+c)=2Q, obviously, P+Q >2Q, according to Theorem 1, know P>Q, so the original inequality is certified. (Note: “=” cannot be used here because the a~2/(a-b)+a-b≥2a, b~2/(b-c)+b-c≥2b equal sign cannot be established at the same time)