In △ ABC, prove: 1 / A + 1 / B + 1 / C > = 9 / A + B + C

In △ ABC, prove: 1 / A + 1 / B + 1 / C > = 9 / A + B + C

My understanding of what you wrote is that a, B and C refer to the side length, and the right side of the inequality is 9 / (a + B + C)
A. Both B and C are greater than 0
(A+B+C)(1/A+1/B+1/C)=3+(A/B+B/A)+(A/C+C/A)+(B/C+C/B)
A/B+B/A>=2((A/B)*(B/A))^1/2=2
Similarly, (A / C + C / a) > = 2, (B / C + C / b) > = 2
Then (a + B + C) (1 / A + 1 / B + 1 / C) > 9

In △ ABC, we prove that: 1 / A + 1 / B + 1 / C > = 9 / π

∵ π / 3 = (a + B + C) / 3 ≥ triple root ABC
1 / cubic root ABC ≥ 3 / π
1 / A + 1 / B + 1 / C ≥ 3 / triple root ABC ≥ 9 / π
OK, I can't get on Baidu