Difficult inequality with 4 variables

Question

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...

Answer 1

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)

Answer 2

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.

Answer 3

Actually the Buffalo Way works. This is outline of a proof.

WLOG, assume that $a = \min(a, b, c, d)$. Let $b=a+u, \ c = a+v$ and $d = a+w$ with $u, v, w\ge 0$. It suffices to prove that $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} + \frac{16abcd}{(a^2+b^2+c^2+d^2)^2} \ge 5$$ or $$q_6 a^6 + q_5a^5 + q_4 a^4 + q_3 a^3 + q_2a^2 + q_1a + q_0 \ge 0$$ where \begin{align} q_6 &= 4(u-v+w)^2, \\ q_5 &= -8(uv-3uw-vw)(u-v+w), \\ q_4 &= 7 u^4+(-24 v+4 w) u^3+(30 v^2-16 v w+54 w^2) u^2\\ &\quad +(-8 v^3-12 v^2 w+16 v w^2+4 w^3) u+7 v^4-24 v^3 w+30 v^2 w^2-8 v w^3+7 w^4,\\ q_3 &= 3 u^5+(-5 v+11 w) u^4+(10 v^2-48 v w+22 w^2) u^3+(6 v^3+22 v^2 w+42 v w^2+22 w^3) u^2\\ &\quad +(7 v^4-48 v^3 w+50 v^2 w^2-24 v w^3+11 w^4) u\\ &\quad +3 v^5-5 v^4 w+10 v^3 w^2+6 v^2 w^3+7 v w^4+3 w^5 ,\\ q_2 &= w^6+(3 u+2 v) w^5+(11 u^2+6 u v+7 v^2) w^4+(6 u^3+2 u v^2) w^3\\ &\quad +(11 u^4+4 u^3 v+46 u^2 v^2+8 u v^3+7 v^4) w^2\\ &\quad +(3 u^5-10 u^4 v-6 u^3 v^2-12 u^2 v^3-9 u v^4-2 v^5) w\\ &\quad +u^6-2 u^5 v+7 u^4 v^2+7 u^2 v^4+2 u v^5+v^6 , \\ q_1 &= (u+v) w^6+(u^2+u v) w^5+(2 u^3+6 u^2 v+7 u v^2+2 v^3) w^4\\ &\quad +(2 u^4-2 u^3 v+2 u^2 v^2-2 u v^3) w^3\\ &\quad +(u^5+5 u^4 v+8 u^3 v^2+6 u^2 v^3+7 u v^4+v^5) w^2\\ &\quad +(u^6-3 u^5 v+2 u^4 v^2-6 u^3 v^3+u^2 v^4-3 u v^5) w\\ &\quad +u^5 v^2+2 u^3 v^4+u v^6 , \\ q_0 &= u^5 v w^2+2 u^3 v^3 w^2+2 u^3 v w^4+u v^5 w^2+2 u v^3 w^4+u v w^6. \end{align} Clearly, $q_6 \ge 0$ and $q_0\ge 0$. We can use Mathematica Resolve to verify $q_4, q_3, q_2, q_1\ge 0$. I hope to see nice proof of them.

Then, it suffices to prove that $q_6 a^6 + q_5a^5 + q_4 a^4 + q_3 a^3 \ge 0$ or $F(a) = q_6a^3 + q_5a^2 + q_4a + q_3 \ge 0$. We can use Mathematica Resolve to verify it. I hope to see nice proof of it.

We are done.