# Difficult inequality with 4 variables

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...

• 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. – Saad Mar 20 at 15:34
• 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. – Olympiados Mar 20 at 15:42
• that is not true as $abcd \leq 1$ – Olympiados Mar 20 at 15:47
• "that is not true as $abcd \leq 1$" what is not true? Please clarify. – астон вілла олоф мэллбэрг Mar 20 at 15:49
• Sorry someone deleted his post...he used AM-GM and the fact that $abcd\geq1$ – Olympiados Mar 20 at 15:51

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)

• Explain please, how from your last inequality follows the original inequality? – Michael Rozenberg Mar 20 at 21:14
• 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 . – ADITYA PRAKASH Mar 21 at 3:18
• 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. – Michael Rozenberg Mar 21 at 5:18
• Actually , the last point is wrong indeed , thanks for pointing it out .Should I delete this answer and post a new one now ? – ADITYA PRAKASH Mar 21 at 6:00
• Yes, of course! – Michael Rozenberg Mar 21 at 6:07

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.

• but if the expression is negative then the inequality is wrong... – Olympiados Mar 20 at 22:00
• @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. – Michael Rozenberg Mar 20 at 22:01
• what do you mean but that? – Olympiados Mar 20 at 22:20
• @Olympiados I did not understand your last question. – Michael Rozenberg Mar 20 at 22:22
• $(u-v+w)^2$ and $(u-v+w)$ , which says this inequality is very strong...what does it mean? – Olympiados Mar 20 at 22:27