Inequality

We can begin by applying the Cauchy-Schwarz inequality to the three pairs of terms under the square roots:

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

Adding these three inequalities, we get:

√(a^2 + ab + b^2) + √(a^2 + ac + c^2) + √(b^2 + bc + c^2) ≥ (a^2 + ab + ac + b^2 + bc + c^2 + ac + bc + c^2)^(1/2)

Simplifying the right-hand side, we get:

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

Now, we can use the AM-GM inequality on the last three terms inside the square root:

2ab + 2ac + 2bc ≥ 3(2(abacbc)^(1/3)) = 3√(a^2b^2c^2)

Substituting this into the previous equation, we get:

(a^2 + b^2 + c^2 + 2ab + 2ac + 2bc)^(1/2) ≥ (a^2 + b^2 + c^2 + 3√(a^2b^2c^2))^(1/2)

Now, we can use the AM-QM inequality on a^2, b^2, and c^2:

(a^2 + b^2 + c^2)/3 ≥ (abc)^(2/3)

Substituting this into the previous equation, we get:

(a^2 + b^2 + c^2 + 3√(a^2b^2c^2))^(1/2) ≥ (4/3)(a^2 + b^2 + c^2)^(1/2)

Therefore, we have:

√(a^2 + ab + b^2) + √(a^2 + ac + c^2) + √(b^2 + bc + c^2) ≥ (4/3)(a^2 + b^2 + c^2)^(1/2)

Finally, we can use the fact that (a^2 + b^2 + c^2) ≥ ab + ac + bc (which follows from the inequality (a-b)^2 + (b-c)^2 + (c-a)^2 ≥ 0) to get:

(4/3)(a^2 + b^2 + c^2)^(1/2) ≥ (4/3)(ab + ac + bc)^(1/2) ≥ (4/3)(3abc)^(1/2) = 2(abc)^(1/2)

Therefore, we have:

√(a^2 + ab + b^2) + √(a^2 + ac + c^2) + √(b^2 + bc + c^2) ≥ 2(abc)^(1/2) = √(4(ab)(ac)(bc)) = √(3(ab) + 3(ac) + 3(bc))

Dividing both sides by √(ab) + √(ac) + √(bc), we get:

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

Therefore, we have shown that

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

Equality occurs when a, b, and c are equal, or when two of them are equal and the third is zero.