Inequality

     
https://math.stackexchange.com/questions/2353736/for-real-a-b-c-if-a2b2c2-abbcac-then-the-value-of-fracab-c
Hint: a^2+b^2+c^2=ab+bc+ca Leftrightarrow Leftrightarrow 2a^2+2b^2+2c^2=2ab+2bc+2ca Leftrightarrow Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0 Leftrightarrow Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0 Leftrightarrow a=b=c
https://math.stackexchange.com/questions/491827/is-this-correct-method-to-prove-that-a2-b2-c2-%E2%89%A5-ab-bc-ac-when-a-b
eginalign & a^2+b^2+c^2-ab-bc-cageq0 \ &iff frac12(2a^2+2b^2+2c^2-2ab-2bc-2ca)geq0 \ &iff frac12((a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2))geq0\ &iff ...

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https://www.quora.com/How-can-I-prove-a-2+b-2+c-2-ab-bc-ca-is-non-negative-for-all-values-of-a-b-and-c
to prove this equation non-negative, you will have to lớn convert the equation in terms of perfect square size containing a,b and c. Now, a²+b²+c²-ab-bc-ca = ½ • ( 2a²+2b²+2c²-2ab-2bc -2ca ) = ½ • ( ...

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https://math.stackexchange.com/questions/1441995/three-line-equations-with-cyclically-permuted-coefficients
If a+b+c=0 và a^2+b^2+c^2=ab+bc+ac, then it follows that 0=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac), or a^2+b^2+c^2=-2(ab+bc+ac). Put this together and we will see that in fact a^2+b^2+c^2=0, ...

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https://math.stackexchange.com/questions/607719/infinite-expansion-of-non-linear-expressions-with-3-or-more-variables/607724
There is an important principle in algebra, that the order of operations must be observed. (a+b+c)^2 requires you to perform the additions first, followed by the squaring operation. A^2+b^2+c^2 ...
HINT: simplifying at first & you will get a^2+ba+ac+b^2+bc+c^2 and now you can plug in the given values
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