$a\sqrt{b} + b\sqrt{c} + c\sqrt{d} + d\sqrt{a} \le 4$

Question

I've got stuck at this problem which I found some days ago in a book about inequalities.

If $a, b, c, d ∈ [0, +\infty)$ and $a+b+c+d=4$, then prove $$a\sqrt{b} + b\sqrt{c} + c\sqrt{d} + d\sqrt{a} \le 4$$

I thought about Cauchy-Buniakowsky-Schwartz inequality but it didn't worked: $$(a^2 + b^2 + c^2 + d^2)(b+c+d+a) \ge (a\sqrt{b} + b\sqrt{c} + c\sqrt{d} + d\sqrt{a})^2$$

I'm looking for some hints, not the entire solution. Thanks!

Answer 1

Because by AM-GM $\sum\limits_{cyc}a\sqrt{b}\leq\sum\limits_{cyc}\frac{a+ab}{2}=2+\frac{(a+c)(b+d)}{2}\leq4$.