For x₁,x₂ > 1, let a = eˣ¹ and b = eˣ², so a,b > e > 2. Then $e^{x_1+x_2}=ab$ and $ab-a-b=(a-1)(b-1)-1>0$. Thus eˣ is superadditive, A. The square root is subadditive for positive inputs; for positive x₁,x₂ the reciprocal of their sum is below either reciprocal, and the decreasing exponential also fails the required inequality.
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