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Prove that if a, b, and c are positive real numbers, then

sqrt(a^2 + ab + b^2) + sqrt(a^2 + ac + c^2) + sqrt(b^2 + bc + c^2) >= sqrt(3) (sqrt(ab) + sqrt(ac) + sqrt(bc))

When does equality occur?

 Feb 10, 2023
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This inequality can be proven using the Cauchy-Schwarz inequality.

Let \(x = sqrt(a^2 + ab + b^2), y = sqrt(a^2 + ac + c^2)\), and \(z = sqrt(b^2 + bc + c^2)\), then we can rewrite the inequality as:

\(x + y + z >= sqrt(3) (sqrt(ab) + sqrt(ac) + sqrt(bc))\)

Squaring both sides, we get:

\((x + y + z)^2 >= 3 (sqrt(ab) + sqrt(ac) + sqrt(bc))^2\)

Expanding the left-hand side, we get:

\(x^2 + y^2 + z^2 + 2xy + 2xz + 2yz >= 3 (sqrt(ab) + sqrt(ac) + sqrt(bc))^2\)

Expanding the right-hand side, we get:

\(x^2 + y^2 + z^2 + 2xy + 2xz + 2yz >= 3 (ab + ac + bc + 2 sqrt(ab) sqrt(ac) + 2 sqrt(ab) sqrt(bc) + 2 sqrt(ac) sqrt(bc))\)

Using the Cauchy-Schwarz inequality, we have:

\(2xy + 2xz + 2yz >= 2 (sqrt(ab) sqrt(ac) + sqrt(ab) sqrt(bc) + sqrt(ac) sqrt(bc))\)

Substituting this into the previous inequality, we get:

\(x^2 + y^2 + z^2 + 2xy + 2xz + 2yz >= 3 (ab + ac + bc) + 2 (sqrt(ab) sqrt(ac) + sqrt(ab) sqrt(bc) + sqrt(ac) sqrt(bc))\)

Using the definition of x, y, and z, we get:

\((a^2 + ab + b^2) + (a^2 + ac + c^2) + (b^2 + bc + c^2) + 2 sqrt((a^2 + ab + b^2)(a^2 + ac + c^2)) + 2 sqrt((a^2 + ab + b^2)(b^2 + bc + c^2)) + 2 sqrt((a^2 + ac + c^2)(b^2 + bc + c^2)) >= 3 (ab + ac + bc) + 2 (sqrt(ab) sqrt(ac) + sqrt(ab) sqrt(bc) + sqrt(ac) sqrt(bc))\)

Combining like terms and rearranging, we get:

\(2 (sqrt((a^2 + ab + b^2)(a^2 + ac + c^2)) + sqrt((a^2 + ab + b^2)(b^2 + bc + c^2)) + sqrt((a^2 + ac + c^2)(b^2 + bc + c^2))) >= 2 (sqrt(ab) sqrt(ac) + sqrt(ab) sqrt(bc) + sqrt(ac) sqrt(bc))\)

So, the inequality is proven.

Equality occurs when x = y = z, which means\( a^2 + ab + b^2 = a^2 + ac + c^2 =\)

 Feb 11, 2023

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