How to prove this inequality where $a$, $b$, $c$, $d$ are positive reals?

Question

Let $a,b,c,d ∈ \Bbb R^+$ , such that $a+b+c+d+abcd = 5$. Prove the inequality:

$1/a+1/b+1/c+1/d \geqslant 4$.

Answer 1

Since $$a+b+c+d \geq 4\sqrt[4]{abcd}$$ we have $$t^4+4t-5\leq 0$$ where $t=\sqrt[4]{abcd}>0$. Becasue of $$(t-1)(t^3+t^2+t+5)\le 0$$ we have $t\leq 1$

Finaly we have: $$1/a+1/b+1/c+1/d \geq 4\sqrt[4]{1/t}\geq 4$$

Answer 2

By $AM-GM$ we get $$\frac{a+b+c+d+abcd}{5}\geq\sqrt[5]{(abcd)^2}$$ so $$1\geq abcd$$ $$AM-GM$$ again $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\geq 4\sqrt[4]{\frac{1}{abcd}}$$ Now we get $$4\sqrt[4]\frac{1}{abcd}\geq 4$$ if $$1\geq abcd$$ and this follows from the condition.