# 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:

\$a^2+b^2+c^2+ab+bc+cage6\$  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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