已知xy为任意实数,求证x^4+y^4≥1/2xy(x+y)^2

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已知xy为任意实数,求证x^4+y^4≥1/2xy(x+y)^2已知xy为任意实数,求证x^4+y^4≥1/2xy(x+y)^2已知xy为任意实数,求证x^4+y^4≥1/2xy(x+y)^2【分析法

已知xy为任意实数,求证x^4+y^4≥1/2xy(x+y)^2
已知xy为任意实数,求证x^4+y^4≥1/2xy(x+y)^2

已知xy为任意实数,求证x^4+y^4≥1/2xy(x+y)^2
【分析法倒推】x^4+y^4≥(1/2)xy(x+y)².2(x^4+y^4)≥xy(x²+2xy+y²)=x³y+xy³+2x²y².(x^4-x³y)+(y^4-xy³)+(x^4-2x²y²+y^4)≥0.x³(x-y)-y³(x-y)+(x²-y²)²≥0.又x³-y³=(x-y)(x²+xy+y²)=(x-y){[x+(y/2)]²+(3y²/4)},故x³(x-y)-y³(x-y)=(x-y)(x³-y³)=(x-y)²{[x+(y/2)]²+(3y²/4)}≥0.等号仅当x=y时取得.又(x²-y²)²≥0.故两非负式相加,也是非负,倒推即得原不等式.