<s>[INST] If $a+b=2$ and $a\geq b>0$ so $a^a(b+1)^2\leq b^b(a+1)^2$ 
Let $a+b=2$, where $a\geq b>0$. Prove that:
  $$a^a(b+1)^2\leq b^b(a+1)^2$$

I tried to prove that $\frac{a^a}{(a+1)^2}\leq\frac{b^b}{(b+1)^2}$, but without success. 
 [/INST] With $a = 1+x, b = 1-x$ we have to show that
$$
 \frac{(1+x)^{1+x}}{(2+x)^2} \le \frac{(1-x)^{1-x}}{(2-x)^2}
 \quad \text{for } 0 \le x < 1 \, .
$$
Taking logarithms this is equivalent to
$$ \tag{*}
 (1+x)\log(1+x) - (1-x)\log(1-x) - 2  \log(2+x) +2 \log(2-x)  \le 0\, .
$$
Denoting the left-hand side with $h(x)$ we have $h(0) = 0$ and
$$
h'(x) = \log(1+x) + \log(1-x) + 2 -  \frac{2}{2+x} - \frac{2}{2-x} \\
   = \log(1-x^2) - \frac{2x^2}{4 - x^2} < 0
$$
for $0 < x < 1$, which implies that $h$ is strictly decreasing.
This completes the proof of $(*)$.
</s>