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(a+b)/2≥ab~(1/2)2/(1/a+1/b)(a>0,b>0)是平均数不等式——“算术平均最大,几何平均次之,调合平均最小”的最简单的情形。它有许多证法,在此介绍一个几何证法作圆,其圆心为A,从圆外一点O引切交OG,切点为G,OA的连线交圆于B、C两点,引GH⊥OB,垂足为H(如图) 令 OC=a,OB=b,则 OA=OC+BC/2=OC+(OB-OC)/2 =(OC+OB)/2=(A+B)/2; ∵ OG~2=OC·OB(切割线定理) ∴ OG=(OC·OB)~(1/2)=ab(1/2); 又 OG~2=OH·OA(射影定理) ∴OH=OG~2/OA=ab/((a+b)/2)=2/(1/a+1/b) 显然,在Rt△OGA中,OA>OG,即(a+b)/2>ab~(1/2);在Rt△OHG中,OG>OH,即ab~(1/2)
(a+b)/2≥ab~(1/2)2/(1/a+1/b) (a>0, b>0) is the mean inequality - "arithmetic mean maximum, geometric mean The simplest case of blending the average minimum. It has many proofs. Here we introduce a geometric proof circle, whose center is A. From a point outside the circle, O leads to OG. The point of tangent is G. The connection of OA is circled at two points, B and C. GH ⊥ OB, pedestal is H (as shown) Let OC=a, OB=b, then OA=OC+BC/2=OC+(OB-OC)/2=(OC+OB)/2=(A+ B)/2; ∵ OG~2=OC·OB (cutting line theorem) ∴ OG=(OC·OB)~(1/2)=ab(1/2); again OG~2=OH·OA (projection) Theorem) ∴OH=OG~2/OA=ab/((a+b)/2)=2/(1/a+1/b) Obviously, in RtΔOGA, OA>OG, ie (a+) b)/2>ab~(1/2); in RtΔOHG, OG>OH, ie ab~(1/2)