If $y = \frac{1}{1+x^{b-a}+x^{c-a}} + \frac{1}{1+x^{c-b}+x^{a-b}} + \frac{1}{1+x^{a-c}+x^{b-c}}$ then $\frac{d^2y}{dx^2}$ is

Take the first fraction: $\frac{1}{1+x^{b-a}+x^{c-a}}$

Multiply both numerator and denominator by $x^a$:

$\frac{1}{1+x^{b-a}+x^{c-a}} = \frac{x^a}{x^a(1+x^{b-a}+x^{c-a})}$

$= \frac{x^a}{x^a + x^{a+b-a} + x^{a+c-a}}$

$= \frac{x^a}{x^a+x^b+x^c}$

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Take the second fraction: $\frac{1}{1+x^{c-b}+x^{a-b}}$

Multiply both numerator and denominator by $x^b$:

$\frac{1}{1+x^{c-b}+x^{a-b}} = \frac{x^b}{x^b(1+x^{c-b}+x^{a-b})}$

$= \frac{x^b}{x^b+x^c+x^a}$

$= \frac{x^b}{x^a+x^b+x^c}$

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Take the third fraction: $\frac{1}{1+x^{a-c}+x^{b-c}}$

Multiply both numerator and denominator by $x^c$:

$\frac{1}{1+x^{a-c}+x^{b-c}} = \frac{x^c}{x^c(1+x^{a-c}+x^{b-c})}$

$= \frac{x^c}{x^c+x^a+x^b}$

$= \frac{x^c}{x^a+x^b+x^c}$

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All three fractions have the same denominator:

$y = \frac{x^a}{x^a+x^b+x^c} + \frac{x^b}{x^a+x^b+x^c} + \frac{x^c}{x^a+x^b+x^c}$

$= \frac{x^a + x^b + x^c}{x^a+x^b+x^c}$

$= 1$

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Since $y = 1$ is a constant:

$\frac{dy}{dx} = 0$

$\frac{d^2y}{dx^2} = 0$

Therefore, $\frac{d^2y}{dx^2} = 0$