# problem regarding application of Jensen's inequality

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

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!
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 .