Inequality $\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$+$\frac{1}{d}$+$\frac{9}{a+b+c+d}\geq 25/4$ 
Prove that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{9}{a+b+c+d}\geq \frac{25}{4}$$
given that $a, b, c, d > 0$ and $abcd = 1$

I reach to a point that says $\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$+$\frac{1}{d}$+$\frac{9}{a+b+c+d}\geq \frac{25}{a+b+c+d}$

Lagrange multipliers.
Let $f(a,b,c,d,\lambda) = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{a+b+c+d} + \lambda(abcd - 1)$. Then for a minimum we have
$$\frac{\partial f}{\partial a} = \frac{\lambda}{a} - \frac{1}{a^2} - \frac{9}{(a+b+c+d)^2} = 0,$$
so $\lambda = \frac{1}{a}+\frac{9a}{(a+b+c+d)^2}$. Since the equation is symmetric, we get
$$\lambda = \frac{1}{a} + \frac{9a}{(a+b+c+d)^2} = \frac{1}{b} + \frac{9b}{(a+b+c+d)^2} = \frac{1}{c} + \frac{9c}{(a+b+c+d)^2} = \frac{1}{d} + \frac{9d}{(a+b+c+d)^2}.$$
From this it follows for each pair of variables that they are the same or their product is $\frac{(a+b+c+d)^2}{9}$. In particular we only have to check cases where $a=b$, $c=d$; $a=b=c$; or $a=b=c=d$. All of these are fairly simple.
It remains to check the boundary. By multiplying by $(a+b+c+d)abcd$ we get an equivalent inequality, which holds also whenever one of the variables is $0$. Notice that if one of the variables approaches $\infty$, some other variable must approach $0$, so we are done.
