Cho `a+b+c=1`. CMR: `a^2+b^2+c^2 ≥ 1/3`

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Ta sẽ chứng minh : `a^2+b^2+c^2>=ab+bc+ca`

`<=>2a^2+2b^2+2c^2-2ab-2bc-2ca>=0`

`<=>(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)>=0`

`<=>(a-b)^2+(b-c)^2+(c-a)^2>=0` (Luôn đúng)

Dấu "`=`" xảy ra khi :

`(a-b)^2=0,(b-c)^2=0,(c-a)^2=0`

`<=>a-b=0,b-c=0,c-a=0`

`<=>a=b=c`

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`a^2+b^2+c^2 >= ab + bc +  ca`

`=> 2a^2+2b^2+ c^2>= 2ab+2bc+2ca`

`=>3a^2+3b^2+3c^2 >= a^2+b^2+c^2+2ab+2bc+2ca`

`=> 3(a^2+b^2+c^2)>= (a+b+c)^2=1^2=1`

`=> a^2+b^2+c^2>= 1/3`

Dấu "`=`" xảy ra khi :

`a=b=c`

Mà `a+b+c=1`

`=>a=b=c=1/3`