If A2+B2+C2 Ab Bc Ca=0 Then Prove That A=B=C

Given that $a$, $b$, $c$ are non-negative real numbers such that $a+b+c=3$, how can we prove that:




By squaring $a+b+c=3$ we get$$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)=9.$$

From the AM-GM inequality (or from the fact that $(x-y)^2=x^2+y^2-2xyge 0$, i.e. $2xyle x^2+y^2$)we have$$ab+ac+bc le fraca^2+b^22+fraca^2+c^22+fracb^2+c^22=a^2+b^2+c^2,$$i.e. $frac12(a^2+b^2+c^2) ge frac12(ab+ac+bc)$, which is equivalent to$frac12(a^2+b^2+c^2) - frac12(ab+ac+bc) ge0$.

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By adding the above equality & inequality together you get$$frac32(a^2+b^2+c^2+ab+ac+bc)ge9,$$which is equivalent to$$a^2+b^2+c^2+ab+ac+bcge6.$$


By the Cauchy-Schwarz inequality,$$6=1*(a+b)+1*(b+c)+1*(a+c)leqsqrt1^2+1^2+1^2sqrt(a+b)^2+(b+c)^2+(a+c)^2$$In other words$$(a+b)^2+(b+c)^2+(a+c)^2geq 12$$which is the same as the desired inequality.



If $x,y,z$ are nonnegative reals, then$x^2+y^2+z^2ge xy+yz+zx$ (with equality iff $x=y=z$), hence$3(x^2+y^2+z^2)ge x(x+y+z)+y(y+z+x)+z(z+x+y) = (x+y+z)^2$(with equality iff $x=y=z$).Letting $x=a+b, y=b+c, z=a+c$, we find $x+y+z=6$ and$3(a+b)^2+3(b+c)^2+3(c+a)^2 ge 36$.Note that $(a+b)^2+(b+c)^2+(c+a)^2= 2(a^2+b^2+c^2+ab+bc+ca)$ so that we actually showed$$ a^2+b^2+c^2+ab+bc+cage 6$$with equality iff $a=b=c$.

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$$ a^2 + b^2 + c^2 + ab + bc + ac = (a+b+c)^2 - (ab + bc + ac) = 9 - (ab + bc + ac)$$Now it remains lớn show that max value of $(ab + bc + ac)$ is $3$. For that, we know the AM-GM equality ( for $a,b, c >0$ ) that $3(a^2 + b^2 + c^2) geq (a+ b + c)^2 geq 3(ab +bc +ac)$. From the last two part we have $(a+ b + c)^2 geq 3(ab +bc +ac) implies 9 geq 3 (ab +bc +ac) implies 3 geq ab +bc +ac$

Hence we have $$ a^2 + b^2 + c^2 + ab + bc + ac = 9 - (ab + bc + ac) geq 9 - 3 = 6$$

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