Question 3
Let $a_1,\ldots,a_n,b_1,\ldots,b_n$ be positive real numbers such that $a_1+\cdots+a_n=b_1+\cdots+b_n$. Show that
$$\frac{a_1^2}{a_1+b_1}+\frac{a_2^2}{a_2+b_2}+\cdots+\frac{a_n^2}{a_n+b_n}\ge\frac{a_1+a_2+\cdots+a_n}{2}.$$
$$\frac{a_1^2}{a_1+b_1}+\frac{a_2^2}{a_2+b_2}+\cdots+\frac{a_n^2}{a_n+b_n}\ge\frac{a_1+a_2+\cdots+a_n}{2}.$$