Cho a, b, c>0. Chứng minh

To prove the inequality, we will use the Cauchy-Schwarz inequality.

By Cauchy-Schwarz inequality, we have:

(ab/c + bc/a + ac/b)(c/a + a/b + b/c) ≥ (√(ab/c * c/a) + √(bc/a * a/b) + √(ac/b * b/c))^2

Simplifying the right side, we get:

(ab/c + bc/a + ac/b)(c/a + a/b + b/c) ≥ (√(b) + √(c) + √(a))^2

(ab/c + bc/a + ac/b)(c/a + a/b + b/c) ≥ (√(a) + √(b) + √(c))^2

Now, let's prove that (√(a) + √(b) + √(c))^2 ≥ 3(a^2 + b^2 + c^2)

Expanding the left side, we get:

(a + b + c + 2√(ab) + 2√(bc) + 2√(ac))^2 ≥ 3(a^2 + b^2 + c^2)

Simplifying, we get:

a^2 + b^2 + c^2 + 4ab + 4bc + 4ac + 4√(ab)√(bc) + 4√(ab)√(ac) + 4√(bc)√(ac) ≥ 3(a^2 + b^2 + c^2)

Rearranging terms, we get:

ab + bc + ac + 2√(ab)√(bc) + 2√(ab)√(ac) + 2√(bc)√(ac) ≥ 2(a^2 + b^2 + c^2)

Using the AM-GM inequality, we have:

ab + bc + ac + 2√(ab)√(bc) + 2√(ab)√(ac) + 2√(bc)√(ac) ≥ 2√(ab * bc) + 2√(ab * ac) + 2√(bc * ac)

Simplifying, we get:

ab + bc + ac + 2√(ab)√(bc) + 2√(ab)√(ac) + 2√(bc)√(ac) ≥ 2√(ab * bc * ac) + 2√(ab * bc * ac) + 2√(ab * bc * ac)

ab + bc + ac + 2√(ab)√(bc) + 2√(ab)√(ac) + 2√(bc)√(ac) ≥ 6√(ab * bc * ac)

Now, let's prove that 6√(ab * bc * ac) ≥ 2(a^2 + b^2 + c^2)

Squaring both sides, we get:

36(ab * bc * ac) ≥ 4(a^2 + b^2 + c^2)^2

9(ab * bc * ac) ≥ (a^2 + b^2 + c^2)^2

Taking the square root of both sides, we get:

3√(ab * bc * ac) ≥ a^2 + b^2 + c^2

Since a, b, c > 0, we can multiply both sides by 3 to get:

9√(ab * bc * ac) ≥ 3(a^2 + b^2 + c^2)

Substituting this inequality back into the previous inequality, we have:

ab + bc + ac + 2√(ab)√(bc) + 2√(ab)√(ac) + 2√(bc)√(ac) ≥ 6√(ab * bc * ac) ≥ 2(a^2 + b^2 + c^2)

Therefore, we have:

(ab/c + bc/a + ac/b)(c/a + a/b + b/c) ≥ (√(a) + √(b) + √(c))^2 ≥ 3(a^2 + b^2 + c^2)

Hence, the inequality ab/c + bc/a + ac/b ≥ √3(a^2 + b^2 + c^2) holds true.