|x＋y|²≤|1＋xy|²,展开,即证：x²＋y²≤1＋x²y²,即证：x²(1－y²)－(1－y²)≤0,即证：(1－y²)(1－x²)≥0.考虑到|x|≤1,|y|≤1,则0≤x²≤1,0≤y²≤...