solve this---A problem from Mexico's Math Olympiad.
Let a, b, c be positive real numbers such that a+b+c=3. Show:
(a^2) / (a + \sqrt[3]{bc}) + (b^2) / (b + \sqrt[3]{ca}) + (c^2) / (c + \sqrt[3]{ab}) >= 3/2
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Today we'll solve a challenging inequality from Mexico's Math Olympiad. We have three positive real numbers a, b, and c with the constraint that their sum equals 3. We need to prove that the sum of three fractions, each with a squared variable in the numerator and a specific expression in the denominator, is at least three-halves. This is a classic example of a cyclic inequality, where the expression has rotational symmetry in the variables. We'll approach this using the powerful Cauchy-Schwarz inequality combined with the AM-GM inequality to establish the required bound.
Let's analyze the key structural properties of this inequality. First, notice the cyclic symmetry - the expression remains unchanged when we cyclically permute the variables a, b, and c. This suggests the minimum occurs at the symmetric point. The constraint a plus b plus c equals 3 is crucial for normalization. The expression has homogeneous structure of degree 1, meaning if we scale all variables by the same factor, the constraint scales accordingly. By symmetry and the method of Lagrange multipliers, we expect equality to occur when a equals b equals c equals 1, which satisfies our constraint. This symmetric point will be our target for verification.
Now we apply the Cauchy-Schwarz inequality, which is our main tool. The Cauchy-Schwarz inequality states that for any vectors u and v, the square of their dot product is less than or equal to the product of their squared magnitudes. We set up our vectors strategically: let u_i equal a_i divided by the square root of the denominator, and v_i equal the square root of the denominator. This gives us that our sum S equals the dot product of u and v. By Cauchy-Schwarz, this is greater than or equal to the square of the sum of u_i, divided by the sum of v_i squared, times the sum of v_i squared. Since the sum of u_i equals a plus b plus c, which is 3, we get S is greater than or equal to 9 divided by the sum of all denominators.
Now we need to analyze the denominator from our Cauchy-Schwarz result. The sum of all denominators equals a plus b plus c, which is 3, plus the sum of the cube roots. We need an upper bound for this sum. By the AM-GM inequality, the sum of cube root of bc, cube root of ca, and cube root of ab is at least 3 times the cube root of their product, which is 3 times the cube root of abc. To bound abc, we again use AM-GM: since a plus b plus c equals 3, we have abc is at most the cube of their average, which is 1. Therefore, the cube root of abc is at most 1, giving us that the sum of denominators is at least 6.
Now we combine our results to complete the proof. From Cauchy-Schwarz, we established that S is greater than or equal to 9 divided by the sum of denominators. From our denominator analysis, we showed this sum is at least 6. Therefore, S is greater than or equal to 9 divided by 6, which equals three-halves. This proves our desired inequality. To verify the equality case, when a equals b equals c equals 1, we have S equals 3 times 1 squared divided by 1 plus cube root of 1, which equals 3 times one-half, giving us exactly three-halves. This confirms that equality occurs at the symmetric point, completing our proof.