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不等式a+b≥2(ab)~(1/2)是中学数学中一个用得很广的基本不等式,但在应用中常见一些错误,现举几例. 一、忽视了a+b≥2(ab)~(1/2)成立条件而导致的错误例1 设a、b、c为正数,求证(a+b+c)~3≥27(a+b-c)(b+c-a)(c+a-b) 错误证法: ∵a+b+c=(a+b-c)+(b+c-a)+(c+a-b)>0 ∴(a+b-c)+(b+c-a)+(c+a-b)≥3((a+b-c)(b+c-a)(c+a-b))~(1/2) 即(a+b+c)~3≥27(a+b-c)(b+c-a)(c+a-b) 分析:虽a>0,b>0,c>0,但a+b-c,b+c-a,c+a-b不一定都大于0,而x+y+z≥3(xyz)~(1/2)的中x、y、z必须都大于0.
The inequality a+b≥2(ab)~(1/2) is a widely used basic inequality in middle school mathematics, but there are some common mistakes in application. Now there are several examples. I. Ignoring a+b≥2 Errors caused by (ab)~(1/2) establishment conditions Example 1 Let a, b, and c be positive numbers, and verify that (a+b+c)~3≥27(a+bc)(b+ca) ( c+ab) wrong proof: ∵a+b+c=(a+bc)+(b+ca)+(c+ab)>0 ∴(a+bc)+(b+ca)+(c+ Ab)≥3((a+bc)(b+ca)(c+ab))~(1/2) (a+b+c)~3≥27(a+bc)(b+ca)( c+ab) analysis: Although a>0, b>0, c>0, but a+bc, b+ca, and c+ab are not always greater than 0, and x+y+z≥3(xyz)~( 1/2) of x, y, z must all be greater than 0.