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Algebra Difficulty 6.0 National olympiad Prove it Austria

Determine the smallest constant CC such that the inequality
(X+Y)2(X2+Y2+C)+(1−XY)2≥0 (X + Y)^2 (X^2 + Y^2 + C) + (1 - XY)^2 \geq 0
holds for all real numbers XX and YY.
For which values of XX and YY does equality hold for this smallest constant CC?

Solution

The smallest constant is C=−1C = -1. Equality holds for X=Y=13X = Y = \frac{1}{\sqrt{3}} or X=Y=−13X = Y = -\frac{1}{\sqrt{3}}.

We first investigate the case X=YX = Y. It is easily seen that the inequality becomes equivalent to
(3X2−1)2+4(C+1)X2≥0 (3X^2 - 1)^2 + 4(C + 1)X^2 \geq 0
which implies C≥−1C \geq -1 by setting X2=13X^2 = \frac{1}{3}.
It remains to prove that the inequality is true for all XX and YY for C=−1C = -1.
Since we had the term (3X2−1)2(3X^2 - 1)^2 in the above case, we compare the term (X2+XY+Y2−1)2(X^2 + XY + Y^2 - 1)^2 with the terms in the given inequality and get the equivalent inequality
(X2+XY+Y2−1)2+(X−Y)2≥0. (X^2 + XY + Y^2 - 1)^2 + (X - Y)^2 \geq 0.
This is obviously true and gives the conditions X=YX = Y and 3X2−1=03X^2 - 1 = 0 for equality which are the two cases X=Y=13X = Y = \frac{1}{\sqrt{3}} and X=Y=−13X = Y = -\frac{1}{\sqrt{3}}.

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