1.令 t =x +y.
因?yàn)?x,y >0,
由基本不等式,
(x +y) /2 ≥√xy,
即 xy ≤(t^2) /4,
當(dāng)且僅當(dāng) x =y =t/2 時(shí)成立.
所以 2 =x +y +xy
≤t +(t^2) /4,
即 t^2 +4t -8 ≥0.
解得 t ≤ -2 -2√3 或 t ≥ 2√3 -2.
所以 x+y ≥ 2√3 -2,
當(dāng)且僅當(dāng) x =y =√3 -1 時(shí),等號成立.
即 當(dāng) x =y =√3 -1 時(shí),x +y 有最小值 2√3 -2.
= = = = = = = = =
換元法,基本不等式,解不等式.
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2.由均值不等式,
√[ (x^2 +y^2) /2 ] ≥ (x +y) /2 ≥ √xy,
當(dāng)且僅當(dāng) x =y 時(shí),等號都成立.
又因?yàn)?x^2 +y^2 =4,
所以 x +y ≤2√(4/2) =2√2,
xy ≤4/2 =2,
當(dāng)且僅當(dāng) x =y =√2 時(shí),等號都成立.
所以 xy +4(x +y) -2 ≤2 +8√2 -2
=8√2.
即 當(dāng) x =y =√2 時(shí),xy +4(x +y) -2 有最小值 8√2.
= = = = = = = = =
均值不等式.
即 平方平均數(shù) ≥算術(shù)平均數(shù) ≥幾何平均數(shù).