#ed_op#br#ed_cl#1/2 = (x+y+z)/3 ≥ #ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√xyz eqv. at x=y=z=1/2 | AM ≥ GM#ed_op#br#ed_cl#=> #ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√[1/(xyz)] ≥ 2 eqv. at x=y=z=1/2#ed_op#br#ed_cl#=> (1/x+1/y+1/z)/3 ≥ #ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√[1/(xyz)] ≥ 2 eqv. at x=y=z=1/2, (1/x)=(1/y)=(1/z)=2 | AM ≥ GM#ed_op#br#ed_cl#=> [(x+1/x)+(y+1/y)+(z+1/z)]/3 = (x+y+z)/3+(1/x+1/y+1/z)/3 ≥ 1/2+2 = 5/2#ed_op#br#ed_cl# eqv. at x=y=z=1/2, (1/x)=(1/y)=(1/z)=2, (x+1/x)=(y+1/y)=(z+1/z)=5/2#ed_op#br#ed_cl#i.e.#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√[(x+1/x)(y+1/y)(z+1/z)] ≥ 3/[1/(x+1/x)+1/(y+1/y)+1/(z+1/z)]#ed_op#br#ed_cl# eqv. at (x+1/x)=(y+1/y)=(z+1/z)=5/2 | GM ≥ HM#ed_op#br#ed_cl# ≥ 3/{[1/(5/2)+1/(5/2)+1/(5/2)]} |(x+1/x=y+1/y=z+1/z=5/2)#ed_op#br#ed_cl# = 5/2#ed_op#br#ed_cl#∴ (x+1/x)(y+1/y)(z+1/z) ≥ (5/2)#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl# = 125/8#ed_op#br#ed_cl#宇智波鼬 寫到:設<math><mrow><mi>x</mi><mo>,</mo><mi>y</mi><mo>,</mo><mi>z</mi><mo>∈</mo><msup><mrow><mi>R</mi></mrow><mrow><mo>+</mo></mrow></msup></mrow></math>#ed_op#br#ed_cl#且<math><mrow><mi>x</mi><mo>+</mo><mi>y</mi><mo>+</mo><mi>z</mi><mo>=</mo><mfrac><mrow><mn>3</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow></math>#ed_op#br#ed_cl#試證:#ed_op#br#ed_cl#<math><mrow><mrow><mo>(</mo><mi>x</mi><mo>+</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mi>x</mi></mrow></mfrac><mo>)</mo><mrow><mo>(</mo><mi>y</mi><mo>+</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mi>y</mi></mrow></mfrac><mo>)</mo><mrow><mo>(</mo><mi>z</mi><mo>+</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mi>z</mi></mrow></mfrac><mo>)</mo><mo>≥</mo><mfrac><mrow><mn>125</mn></mrow><mrow><mn>8</mn></mrow></mfrac></mrow></mrow></mrow></mrow></math>