problem regarding application of Jensen's inequality

Question

question:

For $a,b,c,d \in \mathbb{R^+}$ with $a+b+c+d = 4$, Prove $\displaystyle \sum\dfrac{a}{b(b+1)}\geq \dfrac{8}{(a+c)(b+d)}$

my attempt:

$f(x)= \dfrac{1}{x^2+x}=\dfrac{1}{x(x+1)}$ is convex so, choosing weights $(\omega )$ as $a,b,c,d$ and $x_{i}s$ as $b,c,d,a$ respectively and then applying Jensen's inequality we get;

$ \dfrac{\displaystyle \sum a f(b)}{4}\geq f\left(\dfrac{\displaystyle \sum ab}{4}\right)$

$\displaystyle \sum \dfrac{a}{b(b+1)}\geq 4 f\left(\dfrac{\displaystyle \sum ab}{4}\right)$

$L.H.S\geq 4 \ \dfrac{16}{(\sum ab)[(\sum ab) +4]}= \dfrac{64}{(\sum ab )[( \sum ab)+4]} \neq R.HS = \dfrac{8}{(a+c)(b+d)} $

i'm doing somewhere wrong My left hand side matches with the expression which i have to prove but Right hand side don't match .

so, please provide some hint and answer if possible

thank you

Answer 1

By Holder $$\sum_{cyc}\frac{a}{b(b+1)}\sum_{cyc}ab\sum_{cyc}a(b+1)\geq(a+b+c+d)^3.$$ Thus, since $(a+c)(b+d)=\sum\limits_{cyc}ab$,it's enough to prove that $$(a+b+c+d)^3\geq8\sum_{cyc}(ab+a)$$ or $$\sum_{cyc}ab\leq4,$$ which is true by AM-GM: $$\sum_{cyc}ab=(a+c)(b+d)\leq\left(\frac{a+b+c+d}{2}\right)^2=4$$ and we are done!

Answer 2

You are almost there. Taking from where you left off, observe that $x=ab+bc+cd+da = (a+c)(b+d)$. Thus you need to go one step further to establish $\dfrac{64}{x(x+4)} \ge \dfrac{8}{x}$,and this is clear from AM-GM inequality .