Tintarn wrote: Let $a,b,c$ be real numbers such that $0 \le a \le b \le c$ and $a+b+c=1$. Show that \[ab\sqrt{b-a}+bc\sqrt{c-b}+ac\sqrt{c-a}<\frac{1}{4}.\] Seems that $\sqrt{c-a}(ab+bc+ca) < \frac{1}{5}$ is also true.

This was a wron solution because I thought LHS is cyclic but it is not. However, solution can be repaired and finished

The right answer is shown here: https://youtu.be/wekQLzX6Z5s

By the Schwarz inequality we have, $$\left (ab \sqrt{b-a}+bc\sqrt{c-b}+ca\sqrt{c-a}\right)^2 \leq (ab(b-a)+bc(c-b)+ca(c-a))(ab+bc+ca).$$And so it suffices to prove, $$(ab+bc+ca)(ab(b-a)+bc(c-b)+ca(c-a))<\frac 1{16}(a+b+c)^5.$$We can check this with some computational fortitude: Substitute $(a,b,c)=(x,x+y,x+y+z)$ then take $\text{RHS}-\text{LHS}$ of above. We'll find that all the terms are positive.