Cho x,y,z >0 thỏa mãn x+y+z =4. CMR x+y>= xyz 06/11/2021 Bởi Nevaeh Cho x,y,z >0 thỏa mãn x+y+z =4. CMR x+y>= xyz
16(x+y)=(x+y+z)(x+y+z)(x+y)16(x+y)=(x+y+z)(x+y+z)(x+y) →16(x+y)≥2√(x+y)⋅z⋅2√(x+y)⋅z⋅(x+y)→16(x+y)≥2(x+y)⋅z⋅2(x+y)⋅z⋅(x+y) →16(x+y)≥4(x+y)⋅z⋅(x+y)→16(x+y)≥4(x+y)⋅z⋅(x+y) →16(x+y)≥4(x+y)2⋅z→16(x+y)≥4(x+y)2⋅z →16(x+y)≥4⋅4xy⋅z→16(x+y)≥4⋅4xy⋅z →16(x+y)≥16xyz→16(x+y)≥16xyz →x+y≥xyz→x+y≥xyz Dấu = xảy ra khi x=y,x+y=z→x=y=1/2z Bình luận
16(x+y)=(x+y+z)(x+y+z)(x+y)16(x+y)=(x+y+z)(x+y+z)(x+y)
→16(x+y)≥2√(x+y)⋅z⋅2√(x+y)⋅z⋅(x+y)→16(x+y)≥2(x+y)⋅z⋅2(x+y)⋅z⋅(x+y)
→16(x+y)≥4(x+y)⋅z⋅(x+y)→16(x+y)≥4(x+y)⋅z⋅(x+y)
→16(x+y)≥4(x+y)2⋅z→16(x+y)≥4(x+y)2⋅z
→16(x+y)≥4⋅4xy⋅z→16(x+y)≥4⋅4xy⋅z
→16(x+y)≥16xyz→16(x+y)≥16xyz
→x+y≥xyz→x+y≥xyz
Dấu = xảy ra khi x=y,x+y=z→x=y=1/2z