show that
$$\int_{1}^{e}\dfrac{x^{a+b+c-1}[2(a+b+c)+(c+2a)x^{a-b}+(a+2b)x^{b-c}++(b+2c)x^{c-a}+(2a+b)x^{a-c}+(2b+c)x^{b-a}+(2c+a)x^{c-b}]}{(x^a+x^b)(x^b+x^c)(x^a+x^c)}dx\ge a+b+c$$
and This problem is from this :http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1660742&sid=14f5843656d157b5ce37d5334dc3179d#p1660742
This problem has been bothering me for several years, I have been not been solved. I think this problem maybe use $$(1+x)^n\ge 1+nx,n\in N^*$$
Denote the integral you've written on the left hand side by $I$.
Write $g(x) = (x^a + x^b)(x^a + x^c)(x^b + x^c)$, the denominator. What is $g'(x)$? Try to show that $$ I = \int_1^e \frac{g'(x)}{g(x)} dx $$ so that $$ I = \log g(e) - \log g(1) = \log \left( \frac{1}{8} g(e) \right) $$ Lastly, expand $g(e)$ into a sum of $8$ terms. The function $\log$ is concave downward, which means in particular that for any values $x_1, \cdots, x_8 > 0$, we have $$ \log\left( \frac{1}{8} (x_1 + \cdots + x_8) \right) \geq \frac{1}{8} (x_1 + \cdots + x_8) $$ In this case, the numbers $x_i$ are each of the $8$ pieces you get when expanding $g(e)$. This yields the desired inequality.
