prove that a^2+b^2+c^2-ab-bc-ca is always non-negative for all values of a,b and c
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a^2+b^2+c^2-ab-bc-ca
= 1/2(2a^2+2b^2+2c^2-2ab-2bc-2ca)
= 1/2[(a^2+b^2+c^2)+(a^2+b^2+c^2-2ab-2bc-2ca)]
= 1/2[( a^2+b^2+c^2) + (a-b-c)^2]
For all values of a,b,c;
a^2+b^2+c^2>0
and (a-b-c)^2>0
Therefore 1/2[(a^2+b^2+c^2) + ( a-b-c)^2>0
therefore a^2+b^2+c^2-ab-bc-ca is always positive.
Hence proved!
= 1/2(2a^2+2b^2+2c^2-2ab-2bc-2ca)
= 1/2[(a^2+b^2+c^2)+(a^2+b^2+c^2-2ab-2bc-2ca)]
= 1/2[( a^2+b^2+c^2) + (a-b-c)^2]
For all values of a,b,c;
a^2+b^2+c^2>0
and (a-b-c)^2>0
Therefore 1/2[(a^2+b^2+c^2) + ( a-b-c)^2>0
therefore a^2+b^2+c^2-ab-bc-ca is always positive.
Hence proved!
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Answered by
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Answer:
a^2+b^2+c^2-ab-bc-ca
= 1/2(2a^2+2b^2+2c^2-2ab-2bc-2ca)
= 1/2[(a^2+b^2+c^2)+(a^2+b^2+c^2-2ab-2bc-2ca)]
= 1/2[( a^2+b^2+c^2) + (a-b-c)^2]
For all values of a,b,c;
a^2+b^2+c^2>0
and (a-b-c)^2>0
Therefore 1/2[(a^2+b^2+c^2) + ( a-b-c)^2>0
therefore a^2+b^2+c^2-ab-bc-ca is always positive.
Hence proved!
Step-by-step explanation:
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