If $a+b=2$ so $2+\frac{ab(a-b)^2}{4}\leq a^{1+\sqrt{b}}+b^{1+\sqrt{a}}\leq2+\frac{(a-b)^2}{4}$ 
Let $a$ and $b$ be non-negative numbers such that $a+b=2$. Prove that:
  $$2+\frac{ab(a-b)^2}{4}\leq a^{1+\sqrt{b}}+b^{1+\sqrt{a}}\leq2+\frac{(a-b)^2}{4}$$

In the left inequality the equality occurs also for $a=2$ and $b=0$, which makes additional problems.
For $b=2-a$ we get inequalities of one variable, but derivatives are very ugly and I think it can't help (I mean here $f'(x)=0$...). 
Since the equality occurs for $a=b=1$, we can try the Taylor series, but I  failed in this way.
Thank you!
 A: (Partial solution) For the right inequality:
WLOG, assume $a \le b$. Then $0 < a \le 1 \le b < 2$.
We have $\sqrt{b} = 1 + \frac{b - 1}{\sqrt{b} + 1} \ge 1 + \frac{b - 1}{\frac{b + 1}{2} + 1} = \frac{3b + 1}{b + 3}$
and $\sqrt{a} = 1 - \frac{1 - a}{\sqrt{a} + 1} \le 1 - \frac{1 - a}{(a + 1)/2 + 1} = \frac{3a + 1}{a + 3}$.
It suffices to prove that
$$a^{1 + \frac{3b + 1}{b + 3}} + b^{1 + \frac{3a + 1}{a + 3}}\leq 2+\frac{(a-b)^2}{4}. \tag{1}$$
Let $a = 1 - x, b = 1 + x$ for $0 \le x < 1$. (1) becomes
$$(1 - x)^2 \cdot (1 - x)^{\frac{2x}{4 + x}} + (1 + x)^2 \cdot (1 + x)^{\frac{-2x}{4 - x}} \le 2 + x^2.$$
Fact 1: $(1 + u)^{\frac{-2u}{4 - u}} \le 1 - \frac{1}{2}\,{u}^{2} + \frac{1}{8}\,{u}^{3}-{\frac {1}{96}}\,{u}^{4}+{\frac {11}{384}}
\,{u}^{5} \triangleq F(u)$ for all $u$ in $(-1, 1)$. (Hint: Take logarithm and then take derivatives.)
By Fact 1, it suffices to prove that
$$(1 - x)^2F(-x) + (1 + x)^2F(x) \le 2 + x^2$$
which is written as $\frac{1}{96}x^4(50 - 9x^2) \ge 0$. True.
We are done.
