己知x^２+y^２≤1，,x,y为实数。求证: -√2≤x^２+2xy-y^２≤√2。 证明 令x=rcost,y=rsint,(0≤r≤1),则 |x^２+2xy-y^２|=r^2|(cost)^2+2sintcost-(sint)^2| =r^2|sin2t+cos2t| =√2r^2*|sin(2t+π/4)| ≤√2, 所以 -√2≤x^２+2xy-y^２≤√2.
证明 -√2≤x^２+2xy-y^２≤√2等价于 ︱x^２+2xy-y^２︱/√2≤1 (1) 因为 x^２+y^２≤1，欲证(1)式成立，只需证 ︱x^２+2xy-y^２︱/√2≤x^２+y^２ 即证 -(x^２+y^２)≤(x^２+2xy-y^２)/√2≤(x^２+y^２) (2) 先证(2)式右边，化简整理为: (√2-1)x^２+(√2+1)y^２≥2xy (3) (√2-1)x^２+(√2+1)y^２≥2√[(√2-1)*(√2+1)xy]=2xy. 故(3)成立. 再证(2)式左边化简整理为: (√2+1)x２+(√2-1)y２≥2xy (4) (√2+1)x２+(√2-1)y２≥2√[(√2-1)*(√2+1)xy]=2xy. 故(4)成立.
问：初三代数设X,Y,Z为任意实数,求证:√(X^2+XY+Y^2)+√(X^2+XZ+Z^2)＞√(Y^2+YZ+Z^2)
答：X,Y,Z为任意实数, √(X^2+XY+Y^2)+√(X^2+XZ+Z^2)≥√(Y^2+YZ+Z^2) 仅就Y,Z异号或0给出证明如下(同号时可用类似方法)...详情>>
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