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.
On
Because by Holder $$\sum_{cyc}ab\sum_{cyc}\frac{a}{b(b+1)}=\frac{\sum\limits_{cyc}a(b+1)\sum\limits_{cyc}ab\sum\limits_{cyc}\frac{a}{b(b+1)}}{ab+ac+bc+1}\geq$$ $$\geq\frac{(a+b+c)^3}{ab+ac+bc+1}\geq\frac{(a+b+c)^3}{\frac{(a+b+c)^2}{3}+1}=\frac{3}{4}.$$
On
Remarks: I just remembered that I saw this inequality before. It is a problem of Gabiel Dospinescu. I saw a proof by using integrals.
We have $$\frac{a}{b(b+1)}+\frac{b}{c(c+1)}+\frac{c}{a(a+1)} = \int_0^1 \left(\frac{a}{(x+b)^2} + \frac{b}{(x+c)^2} + \frac{c}{(x+a)^2}\right)\,\mathrm{d} x. \tag{1}$$
By using Cauchy-Bunyakovsky-Schwarz inequality twice, we have $$\frac{a}{(x+b)^2} + \frac{b}{(x+c)^2} + \frac{c}{(x+a)^2} \ge \frac{1}{(x + ab + bc + ca)^2}. \tag{2}$$ (Details are given at the end.)
From (1) and (2), we have $$\frac{a}{b^2 + b} + \frac{b}{c^2 + c} + \frac{c}{a^2 + a} \ge \frac{1}{(ab + bc + ca)(ab + bc + ca + 1)}.$$
Thus, it suffices to prove that $$\frac{1}{ab + bc + ca + 1} \ge \frac34$$ or $$ab + bc + ca \le 1/3$$ which is true using $(a + b + c)^2 \ge 3(ab + bc + ca)$.
We are done.
Details about (2):
By Cauchy-Bunyakovsky-Schwarz inequality, we have $$\left(\frac{a}{(x+b)^2} + \frac{b}{(x+c)^2} + \frac{c}{(x+a)^2}\right)(a + b + c) \ge \left(\frac{a}{x + b} + \frac{b}{x + c} + \frac{c}{x + a}\right)^2$$ and $$\frac{a}{x + b} + \frac{b}{x + c} + \frac{c}{x + a} \ge \frac{(a + b + c)^2}{(ax + ab) + (bx + bc) + (cx + ca)} = \frac{1}{x + ab + bc + ca}.$$ (2) follows.
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.