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Solution
Since 'x' and 'y' are positive integers. Let x = y = 1
x + y = 1 + 1 = 2
$$\sqrt{3xy} = \sqrt{3 * 1 * 1} = \sqrt{3} = 1.732$$
$$\frac{x^2 + y^2}{x + y} = \frac{1^2 + 1^2}{1 + 1} = 1$$
$$\frac{x^4 + x^2y^2 + y^4}{x^3 + y^3} = \frac{1^4 + 1^2 * 1^2 + 1^4}{1^3 + 1^3} = 1.5$$
Hence, x + y is largest.
