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 $\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$+$\frac{1}{d}$+$\frac{9}{a+b+c+d}\geq \frac{25}{a+b+c+d}$
 A: $$\frac{1}{a^4} + \frac{1}{b^4} + \frac{1}{c^4} + \frac{1}{d^4} + \frac{9}{a^4 + b^4 + c^4 + d^4} $$
$$\geq\frac{8}{9}\left(\frac{1}{a^2b^2}+\frac{1}{a^2c^2}+\frac{1}{a^2d^2}+\frac{1}{b^2c^2}+\frac{1}{b^2d^2}+\frac{1}{c^2d^2}\right)+ \frac{11}{3\left(a^4 + b^4 + c^4 + d^4\right)}$$
$$\geq\frac{25}{4abcd},$$
where the last inequality follows from
\begin{align*}
&36a^2b^2c^2d^2(a^4 + b^4 + c^4 + d^4)\left[\dfrac{11}{3t(a^4 + b^4 + c^4 + d^4)}-\frac{25}{4abcd}+\frac{8}{9}\sum_{sym}\frac{1}{a^2b^2}\right]\\
&=\sum_{sym}\dfrac{(a-b)^2}{12}\left((a-b)^2[192(a^2+b^2)(d^2+c^2)+ab(331(c^2-186cd+331d^2)+80cd(c+d)(a+b)
+80c^2d^2]+3cd(c+d)^2(18ab+119c^2-194cd+119d^2)\right)\ge 0
\end{align*}
A: 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.
A: WLOG, assume that $ab \le 1$.
If $ab \le 1/16$, we have $1/a + 1/b \ge 2/\sqrt{ab} \ge 8 > 25/4$.
In the following, assume that $ab > 1/16$.
We split into three cases:
Case 1: $9/16 \le ab \le 1$
Using AM-GM, we have
\begin{align*}
 \mathrm{LHS} &= \left(\frac{1}a + \frac{1}b\right)\left(1 - \frac{9}{16}ab\right)
 + \left(\frac{1}c + \frac{1}d\right)\left(1 - \frac{9}{16}cd\right)\\[5pt]
 &\qquad + \frac{9}{16}(a + b + c + d) + \frac{9}{a + b + c + d}\\[5pt]
 &\ge \frac{2}{\sqrt{ab}}\left(1 - \frac{9}{16}ab\right)
 + \frac{2}{\sqrt{cd}}\left(1 - \frac{9}{16}cd\right) + \frac92\\[5pt]
 &= \frac{2}{\sqrt{ab}}\left(1 - \frac{9}{16}ab\right)
 + 2\sqrt{ab}\left(1 - \frac{9}{16ab}\right) + \frac92\\[5pt]
 &= \frac78\left(\sqrt{ab} + \frac{1}{\sqrt{ab}}\right) + 9/2\\
 &\ge \frac78 \cdot 2 + 9/2\\
 &= 25/4.
\end{align*}
$\phantom{2}$
Case 2: $1/16 < ab < 9/16$ and $a + b \le \frac{1}c + \frac{1}d$
Using AM-GM, we have
\begin{align*}
 \mathrm{LHS} &\ge \frac{1}a + \frac{1}b + \frac{1}c + \frac{1}d + \frac{9}{1/c + 1/d + c + d}\\[5pt]
 &\ge \frac{2}{\sqrt{ab}} + 2\sqrt{\left(\frac{1}c + \frac{1}d\right)\frac{9}{1/c + 1/d + c + d}}\\[5pt]
 &= \frac{2}{\sqrt{ab}} + \frac{6}{\sqrt{1 + cd}}\\
 &= \frac{2}{\sqrt{ab}} + \frac{6\sqrt{ab}}{\sqrt{1 + ab}}.
\end{align*}
Let $x = \sqrt{ab} \in (1/4, 3/4)$.
It suffices to prove that
$$\frac{2}{x} + \frac{6x}{\sqrt{1 + x^2}} \ge 25/4.$$
We have $(1 + x/3)^2 - (1 + x^2)
= \frac29 x (3 - 4x) \ge 0$.
It suffices to prove that
$$\frac{2}{x} + \frac{6x}{1 + x/3} \ge 25/4$$
which is true.
$\phantom{2}$
Case 3: $1/16 < ab < 9/16$ and $a + b > \frac{1}c + \frac{1}d$
From $a + b > 1/c + 1/d = (c + d)/(cd) = (c + d)ab$, we have $1/a + 1/b > c + d$.
Using AM-GM, we have
\begin{align*}
 \mathrm{LHS} &\ge \frac{1}a + \frac{1}b + \frac{1}c + \frac{1}d + \frac{9}{a + b + 1/a + 1/b}\\[5pt]
 &\ge \frac{2}{\sqrt{cd}} + 2\sqrt{\left(\frac{1}a + \frac{1}b\right)\frac{9}{a + b + 1/a + 1/b}}\\[5pt]
 &= \frac{2}{\sqrt{cd}} + \frac{6}{\sqrt{1 + ab}}\\
 &= 2\sqrt{ab} + \frac{6}{\sqrt{1 + ab}}.
\end{align*}
Let $x = \sqrt{ab} \in (1/4, 3/4)$. It suffices to prove that
$$2x + \frac{6}{\sqrt{1 + x^2}} \ge 25/4$$
which is true.
We are done.
A: This is what I wrote first: $a+b+c+d\ge 4$ because $abcd=1$ and arithmetic mean $\ge$ geometric mean. As pointed out by Feanor, this inquality doesn't help from where the OP left.
Alternative: If $a,b,c,d$ are positive, arithmetic mean > geometric mean > harmonic mean so $(a+b+c+d)/4>1>4/(1/a+1/b+1/c+1/d)$. So $1/a+1/b+1/c+1/d+9/(a+b+c+d)>4+9/4=25/4$.
