Question

证明:根号(x^2+xy+y^2)>3/4(x+y)

Answer 1

^2+xy+y^2=(x+y)²-xy
≥(x+y)²-1/4(x+y)²
=3/4(x+y)²
>9/16(x+y)²
根号(x^2+xy+y^2)>3/4(x+y

Answer 2

x^2+xy+y^2
=(x+y)²-xy
≥(x+y)²-1/4(x+y)²
=3/4(x+y)²
>9/16(x+y)²
∴根号(x^2+xy+y^2)>3/4(x+y)

Answer 3

因x²+xy+y²=[x+(y/2)]²+(3y²/4)≥0.故√(x²+xy+y²)有意义。（1）x²+y²≥2xy.===>(x+y)²≥4xy.===>-xy≥-(x+y)²/4.===>x²+xy+y²=(x+y)²-xy≥(x+y)²-[(x+y)²/4]=3(x+y)²/4＞9(x+y)²/16.即x²+xy+y²＞9(x+y)²/16≥0.===>√(x²+xy+y²)＞3|x+y|/4≥(3/4)(x+y).