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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!

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    what is the title of this book? – 2017-01-08
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    "Algebraic Inequalities - from initiation to performance" (title translated, I don't think it's available in English) – 2017-01-08
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    Use your inequality and $$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}} \ge \frac{a+b+c+d}{4}$$ to conclude (this inequality implies $a^2+b^2+c^2+d^2 \ge 4$). – 2017-01-08
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    @Crostul That's not good enough, though. You've proven that the expression is smaller than something which is larger than $4$. That's not what we want. – 2017-01-08
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    @Arthur I see, my bad. – 2017-01-08

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