Difficult inequality with 4 variables I am struggling with this inequality from the book Advanced Olympiad Inequalities: Algebraic & Geometric Olympiad Inequalities, any idea please? Thanks.
Question: Let $a,b,c,d>0$ such that $a^2+b^2+c^2+d^2=4$. Prove that:
$$ \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}+abcd \geq 5 $$
Approach 1: using AM-GM: $$ \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4 $$ but $abcd \leq 1$ so I am not able to conclude.
Approach 2: I have also tried Cauchy-Schwarz, but I am not sure if the inequality that I got is true or not: 
$$ \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}+abcd \geq \frac{(a+b+c+d)^2}{(a+c)(b+d)}+abcd \geq 5$$
But I don't think the last inequality is true...
 A: Tchebychef's inequality-If $a_1,a_2..a_n$ and $b_1,b_2..b_n$ are real numbers then $$\frac{a_1b_1+a_2b_2+ .. +a_nb_n}{n}\geq\left(\frac{a_1+a_1+...+a_n}{n}\right)\left(\frac{b_1+b_1+...+b_n}{n}\right)$$Set $a_i,b_i$ to $a,b,c,d$ .You will quickly get this$$16\geq(a+b+c+d)^2\rightarrow4\geq a+b+c+d$$Now use AM-GM inequality on $a,b,c,d$ to get $$a+b+c+d\geq 4(abcd)^{\frac{1}{4}}$$From the couple of results which have been derived above it is evident that $abcd\leq 1$ . Now applying a final AM-GM inequality.$$\frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}+abcd}{5}\geq(acbd)^{\frac{1}{5}}$$$$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}+abcd\geq5(abcd)^{\frac{1}{5}}$$and we know RHS can never exceed 5 because $abcd\leq1$ , hence the inequality is proven.https://brilliant.org/wiki/chebyshev-inequality/#=  (for Tchebychef's inequality)
A: It's a very long for the comment.
Let $a=\min\{a,b,c,d\}$, $b=a+u$, $c=a+v$ and $d=a+w$.
Thus, we need to prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}+\frac{16abcd}{(a^2+b^2+c^2+d^2)^2}\geq5$$ or
$$4(u-v+w)^2a^6+8(u-v+w)(3uw+vw-uv)a^5+$$
$$+(7u^4+7v^4+7w^4+4u^3w-24u^3v-8v^3u-24v^3w+4w^3u-8w^3v+$$
$$+30u^2v^2+54u^2w^2+30v^2w^2-16u^2vw+12v^2uw-16w^2uv)a^4+...\geq0.$$
The expression, which I wrote can be negative, 
which says that if even there is a proof by BW  so it's very hard. 
