<s>[INST] Prove that $a^ab^bc^cd^d\geq\frac{1}{16}(ab+c+d+1)^2$ 
Given $a, b, c, d\in\mathbb{R^+}$ such that $abcd = 1$. Prove that
$a^ab^bc^cd^d\geq\frac{1}{16}(ab+c+d+1)^2$

I've tried,
$a^ab^bc^cd^d=(abcd)^ab^{b-a}c^{c-a}d^{d-a}=b^{b-a}c^{c-a}d^{d-a}\geq\frac{1}{16}(ab+c+d+1)^2$
i'm stucked
 [/INST] Since $x \mapsto x\ln x$ is convex on $x > 0$, and $a + b + c + d \ge 4\sqrt[4]{abcd} = 4$, we have
\begin{align*}
a\ln a + b\ln b + c\ln c + d\ln d
&\ge 4 \cdot \frac{a + b + c + d}{4} \ln \frac{a + b + c + d}{4}\\
&\ge 4 \ln \frac{a + b + c + d}{4}.
\end{align*}
Also, we have
$$ab + c + d + 1 \le \frac{1}{4}(a + b)^2 + c + d + 1.$$
It suffices to prove that
$$\left(\frac{a + b + c + d}{4}\right)^4 \ge \frac{1}{16}\left(\frac{1}{4}(a + b)^2 + c + d + 1 \right)^2$$
or
$$(a + b + c + d)^2 \ge (a + b)^2 + 4(c + d) + 4$$
or
$$(c + d - 2)^2 + 2(a + b)(c + d) - 8 \ge 0$$
which is true since $2(a + b)(c + d) \ge 2\cdot 2\sqrt{ab} \cdot 2\sqrt{cd} = 8$.
We are done.
</s>