26. (2007.03.30) Simplify
$$\begin{aligned} \sum \frac{a(b+c)}{a^{2}+2 b c}-1-\frac{\sum b c}{\sum a^{2}}= & \sum\left[\frac{a(b+c)}{a^{2}+2 b c}-\frac{a(b+c)}{\sum a^{2}}\right]+ \\ & \frac{2 \sum b c}{\sum a^{2}}-1-\frac{\sum b c}{\sum a^{2}}= \\ & \sum \frac{a(b+c)(b-c)^{2}}{\left(a^{2}+2 b c\right) \sum a^{2}}-\frac{\sum(b-c)^{2}}{2 \sum a^{2}}= \\ & \sum \frac{2 a b+2 a c-a^{2}-2 b c}{2\left(a^{2}+2 b c\right)\left(a^{2}+b^{2}+c^{2}\right)}(b-c)^{2}= \\ & \sum \frac{a^{2}-2(a-b)(a-c)}{2\left(a^{2}+2 b c\right)\left(a^{2}+b^{2}+c^{2}\right)}(b-c)^{2} \end{aligned}$$

Therefore, we only need to prove
$$\sum \frac{a^{2}-2(a-b)(a-c)}{a^{2}+2 b c}(b-c)^{2} \geqslant 0$$

Since
$$\sum \frac{a^{2}(b-c)^{2}}{a^{2}+2 b c} \geqslant 0$$

We only need to prove
$$\begin{array}{l} \sum \frac{-2(a-b)(a-c)(b-c)^{2}}{a^{2}+2 b c} \geqslant 0 \Leftrightarrow \\ \sum \frac{-(a-b)(a-c)(b-c)^{2}}{a^{2}+2 b c} \geqslant 0 \end{array}$$

By symmetry, without loss of generality, assume $a \geqslant b \geqslant c$, then
the left side of equation (3) $=(a-b)(b-c)(a-c)\left(-\frac{b-c}{a^{2}+2 b c}+\frac{a-c}{b^{2}+2 a c}-\frac{a-b}{c^{2}+2 a b}\right)=$
$$\begin{array}{l} (a-b)(b-c)(a-c)\left[(b-c)\left(\frac{1}{b^{2}+2 a c}-\frac{1}{a^{2}+2 b c}\right)+\right. \\ \left.(a-b)\left(\frac{1}{b^{2}+2 a c}-\frac{1}{c^{2}+2 a b}\right)\right]= \\ (a-b)(b-c)(a-c)\left[\frac{(a+b-2 c)(a-b)(b-c)}{\left(b^{2}+2 a c\right)\left(a^{2}+2 b c\right)}+\right. \\ \left.\frac{(2 a-b-c)(a-b)(b-c)}{\left(b^{2}+2 a c\right)\left(c^{2}+2 a b\right)}\right] \geqslant 0 \end{array}$$

Thus, equation (3) holds. By equations (2) and (3), equation (1) holds, and the original proposition is proved.