I'm interested by the following problem :
Let $a,b,c,d>0$ such that $a+b+c+d=4$ then we have : $$ab\ln(a)+bc\ln(b)+cd\ln(c)+da\ln(d)> -1.37$$
My try for $b\geq 3,a\leq1,c\leq 1,d\leq1$:
We have the following inequality :
Let $\varepsilon >0$ and $x>0$ then we have : $$\frac{x(1-\varepsilon)}{\varepsilon}-\frac{(x)^{1-\varepsilon}}{\varepsilon e^{\varepsilon}}\leq x\ln(x)$$
So we have to prove (here I take $\varepsilon=0.001$):
$$\frac{ab(1-\varepsilon)}{\varepsilon}-\frac{b(a)^{1-\varepsilon}}{\varepsilon e^{\varepsilon}}+\frac{bc(1-\varepsilon)}{\varepsilon}-\frac{c(b)^{1-\varepsilon}}{\varepsilon e^{\varepsilon}}+\frac{cd(1-\varepsilon)}{\varepsilon}-\frac{d(c)^{1-\varepsilon}}{\varepsilon e^{\varepsilon}}+\frac{da(1-\varepsilon)}{\varepsilon}-\frac{a(d)^{1-\varepsilon}}{\varepsilon e^{\varepsilon}}>-1.37$$
Now I use a form of Jensen's inequality :
Let $a,b,c,d>0$ such that $a+b+c+d=4$ and $b\geq 3,a\leq1,c\leq 1,d\leq1$ then we have : $$3.7(\frac{ab+bc+cd+da}{3.7})^{1-\varepsilon}>b(a)^{1-\varepsilon}+c(b)^{1-\varepsilon}+d(c)^{1-\varepsilon}+a(d)^{1-\varepsilon}$$
If we combine this two fact we have to prove : $$\frac{(ab+bc+cd+da)(1-\varepsilon)}{\varepsilon}-\frac{3.7(\frac{ab+bc+cd+da}{3.7})^{1-\varepsilon}}{\varepsilon e^{\varepsilon}}>-1.37$$
Wich is a one variable inequality because $$\varepsilon=0.001$$
I have two questions :
How to complete my reasoning ? How to prove this version of Jensen's inequality ?
Thanks in advance for your time .