Prove that $a^2 + b^2 + c^2 – ab – bc – ca$ is always non-negative for all values of $a, b$ and $c$.

To do:

We have to prove that $a^2 + b^2 + c^2 – ab – bc – ca$ is always non-negative for all values of $a, b$ and $c$.

Solution:

We know that,

$(a+b)^2=a^2+b^2+2ab$

$(a-b)^2=a^2+b^2-2ab$

$(a+b)(a-b)=a^2-b^2$

Therefore,

$a^{2}+b^{2}+c^{2}-a b-b c-c a$

Multiplying and dividing the above expression by 2, we get,

$\frac{2}{2}\times a^{2}+b^{2}+c^{2}-a b-b c-c a=\frac{1}{2}[2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 c a]$

$=\frac{1}{2}[a^{2}+b^{2}-2 a b+b^{2}+c^{2}-2 b c+c^{2}+a^{2}-2 c a]$

$=\frac{1}{2}[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]$

The sum of the squares of any numbers is always non-negative for all values of $a, b$ and $c$.

Hence proved.

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Updated on: 10-Oct-2022

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