Problem 1258

Prove the left inequality first.
Square both sides to get
$$\begin{array}{l} \sum \frac{a^{2}}{a+b}+2 \sum \frac{a b}{\sqrt{a+b} \sqrt{b+c}} \\ \geqslant\left(\sum a b\right)\left(\sum \frac{1}{a+b}\right) \\ \Leftrightarrow \sum\left(a-\frac{a b}{a+b}\right)+2 \sum \frac{a b}{\sqrt{a+b} \sqrt{b+c}} \\ \geqslant \sum\left(a+\frac{a b}{a+b}\right) \\ \Leftrightarrow \sum \frac{a b}{\sqrt{a+b} \sqrt{b+c}} \geqslant \sum \frac{a b}{a+b} \\ \Leftrightarrow \sum a b(a+c) \sqrt{(a+b)(b+c)} \\ \quad \geqslant \sum a b(a+c)(b+c) . \end{array}$$

Notice that,
$$\begin{array}{l} (a+b)(b+c)-\left(b+\frac{2 a c}{a+c}\right)^{2} \\ =\frac{a b+b c+c a}{(a+c)^{2}}(a-c)^{2} \geqslant 0 . \\ \text { Hence } a a b(a+c) \sqrt{(a+b)(b+c)} \\ \geqslant \sum a b(a+c)\left(b+\frac{2 a c}{a+c}\right) \\ =\sum a b(a+c)(b+c) . \end{array}$$

Now prove the right inequality.
Using the weighted Jensen's inequality, we get
$$\begin{array}{l} \frac{\sqrt{2}}{2}(\sqrt{a}+\sqrt{b}+\sqrt{c}) \\ =\frac{\sqrt{2}}{2} \times 2\left(\sum \frac{1}{a}\right)\left[\sum \frac{\frac{1}{b}+\frac{1}{c}}{2 \sum \frac{1}{a}} \sqrt{\frac{a}{\left(\frac{1}{b}+\frac{1}{c}\right)^{2}}}\right] \\ \leqslant \frac{\sqrt{2}}{2} \times 2\left(\sum \frac{1}{a}\right) \sqrt{\sum \frac{a}{\left(\frac{1}{b}+\frac{1}{c}\right) 2\left(\sum \frac{1}{a}\right)}} \\ =\frac{\sqrt{2}}{2} \sqrt{\left(\sum \frac{a}{\frac{1}{b}+\frac{1}{c}}\right) 2\left(\sum \frac{1}{a}\right)} \\ =\sqrt{\left(\sum a b\right)\left(\sum \frac{1}{a+b}\right)} \text {. } \\ \end{array}$$

In conclusion, the result holds.