<s>[INST] 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 $\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$+$\frac{1}{d}$+$\frac{9}{a+b+c+d}\geq \frac{25}{a+b+c+d}$
 [/INST] 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 $1/a+1/b+1/c+1/d+9/(a+b+c+d)>4+9/4=25/4$.
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