Question

CMR: a²+b²+c²-ab-bc-ca>=0 với mọi a, b, c

Answer 1

\(a^2\)+\(b^2\)+\(c^2\)-ab-bc-ca\(\ge\)0

<=> 2\(a^2\)+2\(b^2\)+2\(c^2\)-2ab-2bc-2ac\(\ge\)0

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

<=> \(\left(a-b\right)^2\)+\(\left(b-c\right)^2\)+\(\left(c-a\right)^2\)\(\ge\)0

vì \(\left(a-b\right)^2\)\(\ge\)0 

\(\left(b-c\right)^2\)\(\ge\)0

\(\left(c-a\right)^2\)\(\ge\)0

<=>\(\left(a-b\right)^2\)+\(\left(b-c\right)^2\)+\(\left(c-a\right)^2\)\(\ge\)0

vậy\(a^2\)+\(b^2\)+\(c^2\)-ab-bc-ca\(\ge\)0

dấu = xảy ra khi

a-b=0=>a=b

b-c=0=> b=c

c-a=0=> c=a

=> a=b=c

Answer 2

Mày nhìn cái chóa j