1/a^2+b^2+c^2 + 2023/(ab+bc+ca) lớn hơn hoặc 2024/3

To prove the inequality, we can use the AM-GM inequality.

By AM-GM, we have:

ab + bc + ca ≥ 3√(ab * bc * ca)

Since a, b, and c are positive, we can rewrite this as:

ab + bc + ca ≥ 3√(a * b * c)^2

ab + bc + ca ≥ 3abc

Now, let's simplify the left side of the inequality:

1/a^2 + b^2 + c^2 + 2023/(ab + bc + ca)

= (1/a^2 + b^2 + c^2) + 2023/(ab + bc + ca)

Using AM-GM again, we have:

(1/a^2 + b^2 + c^2) + 2023/(ab + bc + ca) ≥ 2√((1/a^2 + b^2 + c^2) * (2023/(ab + bc + ca)))

= 2√((1/a^2 + b^2 + c^2) * (2023/(3abc)))

= 2√((1/a^2 + b^2 + c^2) * (674.333.../abc))

Now, let's simplify the right side of the inequality:

2024/3 = 674.333...

Therefore, we have:

2√((1/a^2 + b^2 + c^2) * (674.333.../abc)) ≥ 674.333...

This shows that:

1/a^2 + b^2 + c^2 + 2023/(ab + bc + ca) ≥ 2024/3

Hence, the inequality is proven.