Selection Examination

Since $a^2+b^2+c^2=3$, the inequality is equivalent to

$$\left(\frac{a^2+b^2}{2ab}+1\right)+\left(\frac{b^2+c^2}{2bc}+1\right)+\left(\frac{c^2+a^2}{2ca}+1\right)+\left(\frac{2(ab+bc+ca)}{a^2+b^2+c^2}+1\right)\ge 9.$$ which can be written as $$\frac{(a+b{)}^2}{2ab}+\frac{(b+c{)}^2}{2bc}+\frac{(c+a{)}^2}{2ca}+\frac{(a+b+c{)}^2}{a^2+b^2+c^2}\ge 9.$$ Let $L$ be the left hand side of the last inequality. Then from Cauchy-Schwarz inequality in the form of Engel or Andreescu it follows that $$L\ge \frac{((a+b)+(b+c)+(c+a)+(a+b+c){)}^2}{2ab+2bc+2ca+a^2+b^2+c^2}=\frac{9(a+b+c{)}^2}{(a+b+c{)}^2}=9.$$ The equality holds, if and only if $$\frac{a+b}{2ab}=\frac{b+c}{2bc}=\frac{c+a}{2ca}=\frac{a+b+c}{3}.$$ Since $a,b,c>0$, from the first two equations we get $a=b=c$, and by using the last equation we find: $a=b=c=1$.