Inequality $\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$+$\frac{1}{d}$+$\frac{9}{a+b+c+d}\geq 25/4$ $\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$+$\frac{1}{d}$+$\frac{9}{a+b+c+d}\geq 25/4$

given than $abcd = 1$

I reach to a point that says $\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$+$\frac{1}{d}$+$\frac{9}{a+b+c+d}\geq \frac{25}{a+b+c+d}$
$$\frac{1}{a^4} + \frac{1}{b^4} + \frac{1}{c^4} + \frac{1}{d^4} + \frac{9}{a^4 + b^4 + c^4 + d^4} $$

$$\geq\frac{8}{9}\left(\frac{1}{a^2b^2}+\frac{1}{a^2c^2}+\frac{1}{a^2d^2}+\frac{1}{b^2c^2}+\frac{1}{b^2d^2}+\frac{1}{c^2d^2}\right)+ \frac{11}{3\left(a^4 + b^4 + c^4 + d^4\right)}$$

$$\geq\frac{25}{4abcd},$$

where the last inequality follows from

\begin{align*}
&amp;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]\\
&amp;=\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*}
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 $(a+b+c+d)/4&gt;1&gt;4/(1/a+1/b+1/c+1/d)$. So $1/a+1/b+1/c+1/d+9/(a+b+c+d)&gt;4+9/4=25/4$.