Cho a+b+c=1. Chứng minh: 1/a^2+b^2+c^2 + 1/a^2 + 1/b^2 + 1/c^2 >= 30

To solve this problem, we can use the Cauchy-Schwarz inequality.

By Cauchy-Schwarz inequality, we have:

(a^2 + b^2 + c^2)(1/a^2 + 1/b^2 + 1/c^2) ≥ (1 + 1 + 1)^2

Simplifying, we get:

(1/a^2 + 1/b^2 + 1/c^2) ≥ 9

Now, we need to find the minimum value of (1/a^2 + 1/b^2 + 1/c^2).

Using AM-HM inequality, we have:

(a^2 + b^2 + c^2)/3 ≥ 3/(1/a^2 + 1/b^2 + 1/c^2)

Simplifying, we get:

(1/a^2 + 1/b^2 + 1/c^2) ≥ 9/(a^2 + b^2 + c^2)

Now, substituting the given condition a + b + c = 1, we have:

(1/a^2 + 1/b^2 + 1/c^2) ≥ 9/(1 - 2ab - 2bc - 2ca)

Since a + b + c = 1, we can rewrite the denominator as:

(1 - 2ab - 2bc - 2ca) = (a + b + c)^2 - 2(ab + bc + ca) = 1 - 2(ab + bc + ca)

Therefore, we have:

(1/a^2 + 1/b^2 + 1/c^2) ≥ 9/(1 - 2(ab + bc + ca))

To find the minimum value of (1/a^2 + 1/b^2 + 1/c^2), we need to find the maximum value of (ab + bc + ca).

By AM-GM inequality, we have:

ab + bc + ca ≤ (a^2 + b^2 + c^2)/2

Substituting the given condition a + b + c = 1, we have:

ab + bc + ca ≤ (1 - a^2 - b^2 - c^2)/2

Since a^2 + b^2 + c^2 ≥ ab + bc + ca, we can rewrite the inequality as:

ab + bc + ca ≤ (1 - (ab + bc + ca))/2

Simplifying, we get:

3(ab + bc + ca) ≤ 1

Therefore, we have:

(1/a^2 + 1/b^2 + 1/c^2) ≥ 9/(1 - 2(ab + bc + ca)) ≥ 9/(1 - 2/3) = 27

Hence, the minimum value of (1/a^2 + 1/b^2 + 1/c^2) is 27.

Therefore, we have:

1/a^2 + b^2 + c^2 + 1/a^2 + 1/b^2 + 1/c^2 ≥ 30

This inequality is satisfied when (1/a^2 + 1/b^2 + 1/c^2) = 27.

Hence, the given inequality is true.