Prove that $(ab+bc+ca)\left( \frac{a}{b(b+1)}+\frac{b}{c(c+1)}+\frac{c}{a(a+1)}\right)\ge \frac{3}{4}$ I would appreciate if somebody could help me with the following problem:
Q: Prove that where $a,b,c \in [0,1]$, $a+b+c=1$
$$(ab+bc+ca)\left( \frac{a}{b(b+1)}+\frac{b}{c(c+1)}+\frac{c}{a(a+1)}\right)\ge
 \frac{3}{4}$$
My work: The first method tried was to use the arithmetic-geometric mean because of the positive number condition. However, the expression of the problem becomes more complicated, but the solution is not visible.

The function $f(x) = \frac{1}{x(x+1)}$ is convex on $x > 0$ (note: $f''(x) = \frac{6x^2 + 6x + 2}{x^3(x+1)^3} > 0$).
Thus, by Jensen's inequality, we have
\begin{align*}
 \frac{a}{b(b+1)}+\frac{b}{c(c+1)}+\frac{c}{a(a+1)} 
 &\ge f(ab + bc + ca)\\
 &= \frac{1}{(ab+bc+ca)(ab+bc+ca+1)}.
\end{align*}
Thus, we have
\begin{align*}
 &(ab+bc+ca)\left( \frac{a}{b(b+1)}+\frac{b}{c(c+1)}+\frac{c}{a(a+1)}\right)\\
 \ge{}& (ab+bc+ca)\cdot \frac{1}{(ab+bc+ca)(ab+bc+ca+1)}\\
 ={}& \frac{1}{ab+bc+ca+1}\\
 \ge{}& \frac{1}{1/3 + 1}\\
 ={}& \frac34
\end{align*}
where we have used $ab + bc + ca \le (a + b + c)^2/3 = 1/3$.
We are done.
