Đáp án:
Giải thích các bước giải:
Ta có:
`(1-x)^2≥0∀x`
`(x-y)^2≥0∀x;y`
`(y-z)^2≥0∀y;z`
`=>(1-x)^2 + (x-y)^2 +(y-z)^2≥0∀x;y;z`
Dấu "=" xáy ra khi: `1-x=x-y=y-z=0⇔x=y=z=1`
⇒x=y=z đpcm
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Giúp mình trả lời câu hỏi này với Phân biệt điểm khác nhau giữa thuyết minh và thuyết trình
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