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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...

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  • $\begingroup$ This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc. $\endgroup$ – Saad Mar 20 at 15:34
  • $\begingroup$ I tried the classic approaches such as AM-GM, Cauchy-Schwarz but they don't work. I haven't tried BW but maybe that is the last option. $\endgroup$ – Olympiados Mar 20 at 15:42
  • $\begingroup$ that is not true as $abcd \leq 1$ $\endgroup$ – Olympiados Mar 20 at 15:47
  • $\begingroup$ "that is not true as $abcd \leq 1$" what is not true? Please clarify. $\endgroup$ – астон вілла олоф мэллбэрг Mar 20 at 15:49
  • $\begingroup$ Sorry someone deleted his post...he used AM-GM and the fact that $abcd\geq1$ $\endgroup$ – Olympiados Mar 20 at 15:51
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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)

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  • $\begingroup$ Explain please, how from your last inequality follows the original inequality? $\endgroup$ – Michael Rozenberg Mar 20 at 21:14
  • $\begingroup$ The LHS is the original inequality and RHS is 5 multiplied by 'something' , and that 'something' can never exceed 1 since fourth root of a quantity less than 1 , never exceeds 1 , overall 5 times 'something' never exceeds 1 , therefore RHS never exceeds 5 . $\endgroup$ – ADITYA PRAKASH Mar 21 at 3:18
  • $\begingroup$ Let $LS=A$, $\sqrt[5]{abcd}=B$ and $1=C$. We need to prove that $A\geq C$. You proved that $A\geq B$ and $C\geq B$. Why $A\geq C$ is true? It can be wrong. For example, $2>0$ and $3>0$, but $2>3$ is wrong. $\endgroup$ – Michael Rozenberg Mar 21 at 5:18
  • $\begingroup$ Actually , the last point is wrong indeed , thanks for pointing it out .Should I delete this answer and post a new one now ? $\endgroup$ – ADITYA PRAKASH Mar 21 at 6:00
  • $\begingroup$ Yes, of course! $\endgroup$ – Michael Rozenberg Mar 21 at 6:07
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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.

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  • $\begingroup$ but if the expression is negative then the inequality is wrong... $\endgroup$ – Olympiados Mar 20 at 22:00
  • $\begingroup$ @Olympiados No, it does not say this because there are expressions else. I think, it's interesting that we got factors $(u-v+w)^2$ and $u-v+w,$ which says that this inequality is very strong. $\endgroup$ – Michael Rozenberg Mar 20 at 22:01
  • $\begingroup$ what do you mean but that? $\endgroup$ – Olympiados Mar 20 at 22:20
  • $\begingroup$ @Olympiados I did not understand your last question. $\endgroup$ – Michael Rozenberg Mar 20 at 22:22
  • $\begingroup$ $(u-v+w)^2$ and $(u-v+w)$ , which says this inequality is very strong...what does it mean? $\endgroup$ – Olympiados Mar 20 at 22:27

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