Problem 1202

We consider first the left hand inequality. As $a + b + c = 1$ we have:
$$a \leq 1 \leq 1 + b, \text{ so } a = \sqrt{a^2} \leq \sqrt{a(1+b)}$$
Applying this similarly to the other square roots and adding quickly gives:
$$1 = a + b + c \leq \sqrt{a(1+b)} + \sqrt{b(1+c)} + \sqrt{c(1+a)}.$$
A direct application of the AM-GM inequality does not yield the right hand inequality. Instead, we can tighten the AM-GM inequality by scaling the factors so they are equal when $a = b = 1/3$. This gives:
$$\sqrt{a(1+b)} = \sqrt{2a \cdot \frac{1+b}{2}} \leq \frac{2a}{2} + \frac{1+b}{4}.$$
We can add the three cyclical permutations and obtain:
$$\sqrt{2a \cdot \frac{1+b}{2}} + \sqrt{2b \cdot \frac{1+c}{2}} + \sqrt{2c \cdot \frac{1+a}{2}} \\ \leq a + \frac{1+b}{4} + b + \frac{1+c}{4} + c + \frac{1+a}{4} = a + b + c + \frac{3 + a + b + c}{4} = 2.$$

Alternatively, to prove the right hand inequality we may use the Cauchy-Schwarz inequality
$$\left| \sum_{k=1}^{n} x_k y_k \right| \leq \sqrt{\sum_{k=1}^{n} x_k^2} \cdot \sqrt{\sum_{k=1}^{n} y_k^2}$$
with $n = 3$, $x_1 = \sqrt{a}$, $x_2 = \sqrt{b}$, $x_3 = \sqrt{c}$ and $y_1 = \sqrt{b+1}$, etc. We obtain
$$\sqrt{a(1+b)} + \sqrt{b(1+c)} + \sqrt{c(1+a)} \\ \leq \sqrt{a + b + c} \cdot \sqrt{(1+b) + (1+c) + (1+a)} = \sqrt{1} \cdot \sqrt{4} = 2.$$