# How prove this inequality $\sum\limits_{cyc}\frac{1}{a+3}-\sum\limits_{cyc}\frac{1}{a+b+c+1}\ge 0$

show that: $$\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}-\left(\dfrac{1}{a+b+c+1}+\dfrac{1}{b+c+d+1}+\dfrac{1}{c+d+a+1}+\dfrac{1}{d+a+b+1}\right)\ge 0$$ where $abcd=1,a,b,c,d>0$

I have show three variable inequality

Let $a$, $b$, $c$ be positive real numbers such that $abc=1$. Prove that $$\frac{1}{1+b+c}+\frac{1}{1+c+a}+\frac{1}{1+a+b}\leq\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$$ also see:can see ：http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=243

from this equality,I have see a nice methods:

I think this Four varible inequality is also true

• The choices $x := 0.1,$ $a := 1,$ $b, c := 1/2$, $d := 4$ seem to constitute a counterexample. – Benicio Aug 24 '14 at 4:20
• Please can you avoid strong $TeX$ in title for the slower computers. – Tony Aug 24 '14 at 5:32
• @Brian, could you please tell us how your $x$ is related to the question? For me $a=1,b=1/2,c=1/2,d=4$ give $57/364$ which I suppose is positive. – andre Aug 30 '14 at 10:07
• did you noticed my answer? – RE60K Aug 30 '14 at 11:05

We have \begin{align} \frac3{a+b+c+1} &- \frac1{a+3} - \frac1{b+3} - \frac1{c+3}\\ &= \sum_{cyc}^{a, b, c} \left(\frac1{a+b+c+1}-\frac1{a+3} \right) \\ &= \frac1{a+b+c+1} \sum_{cyc}^{a, b, c}\frac{2-b-c}{a+3} \\ &= \frac1{(a+b+c+1)\prod_{cyc}^{a, b, c}(a+3)} \sum_{cyc}^{a, b, c} (18-6a-4ab-a^2b-ab^2-6a^2) \\ &\le \frac1{1\times3^3} \sum_{cyc}^{a, b, c} (18-6a-4ab-a^2b-ab^2-6a^2) \end{align}

using $a, b, c > 0$. Summing over four such inequalities, we get $$3\sum_{cyc}\frac1{a+b+c+1} - 3\sum_{cyc} \frac1{a+3} \\ \le \frac2{27}\left( 108-\sum_{cyc}\left(9(a+a^2)+4(2ab+bd)+(a^2b+ab^2+a^2c+ac^2+b^2d+bd^2)\right) \right)$$

Now by AM-GM and the constraint, we have that $\sum_{cyc} a^mb^n \ge 4\sqrt[4]{(abcd)^{m+n}}=4$ for all $m, n \ge 0$, so RHS $\le 0$ and we are done.

P.S. the method looks general, though I wouldn't want to write down the cyclic sums for more variables!

• Nice solution! I may well be wrong about this, but I think the RHS of the last inequality should be $\frac{1}{27}[\sum_{cyc}^{abcd}(54-18a-8ab-2a^2b-2ab^2-18a^2)-8(ac+bd)-2(a^{2}c+ac^2+bd^2+b^{2}d)]$, but your argument is still valid even if this is the case. – user84413 Sep 4 '14 at 16:33
• Again, I may be wrong, but I think the 144 should be 108 instead, and also that the terms $a^2c,ac^2,b^2d,bd^2$ are getting counted 6 times instead of twice. – user84413 Sep 5 '14 at 22:35
• Also, I think perhaps the terms $ac,bd$ are getting counted 32 times instead of 8 times. – user84413 Sep 5 '14 at 22:54
• Okay, but I think the 54 you had originally was correct, so I'm not sure how you're getting 144 (since $2\times144\ne4\times54$). – user84413 Sep 6 '14 at 22:33

Let's use Lagrange Multipliers: Let use take $f$ to be: $$f(a,b,c,d)=\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}-\left(\dfrac{1}{a+b+c+1}+\dfrac{1}{b+c+d+1}+\dfrac{1}{c+d+a+1}+\dfrac{1}{d+a+b+1}\right)$$ subject to constraint $abcd=1$: $$g(a,b,c,d)=abcd=1$$ Here are the equations we need to solve: $$\frac{-1}{(a+3)^2}+\sum_{cyc}\dfrac{1}{(a+b+c+1)^2}=\lambda bcd\\ \frac{-1}{(b+3)^2}+\sum_{cyc}\dfrac{1}{(a+b+c+1)^2}=\lambda acd\\ \frac{-1}{(c+3)^2}+\sum_{cyc}\dfrac{1}{(a+b+c+1)^2}=\lambda abd\\ \frac{-1}{(d+3)^2}+\sum_{cyc}\dfrac{1}{(a+b+c+1)^2}=\lambda abc\\ abcd=1$$ Let $\displaystyle Z:=\sum_{cyc}\dfrac{1}{(a+b+c+1)^2}$ $$Z=\frac{\lambda}a+\frac1{(a+3)^2}=\frac{\lambda}b+\frac1{(b+3)^2}=\frac{\lambda}c+\frac1{(c+3)^2}=\frac{\lambda}d+\frac1{(d+3)^2}$$ Since $a=b=c=d$ seems to satisfy the equations from intuition. If you don't believe see this page on W|A, which says that when $ab(a+3)(b+3)\ne0$(which is true as $a,b,c,d>0$) the solutions(solved for two variables) are: $$a=b,\lambda=-\frac{(a b (a+b+6))}{((a+3)^2 (b+3)^2)}$$ Since, now $a=b=c=d(=1)$, the value of $f$ seems to be: $$f(1,1,1,1)=4\times\frac14-4\times\frac14=0$$ Which apparently is its minimum value.

Now for the general case of n-variables the equations would be: $$\frac{-1}{(x_1+(n-1))^2}+\sum_{cyc}\dfrac{1}{(\sum_{cyc}x_i+1)^2}=\lambda \frac{\prod x_i}{x_1}=\frac{\lambda}x_1\tag{cycle equation 'n' times}\\\prod x_i=1$$ Again let $\displaystyle Z:=\sum_{cyc}\dfrac{1}{(\sum_{cyc}x_i+1)^2}$ Now $$Z=\frac{\lambda}x_i+\frac{1}{(x_1+(n-1))^2}$$ Again extending that logic we would get $$x_1=x_2=\cdots=x_n(=1)$$ So the value of $f$ here would be: $$f(1,1,\cdots,1)=\sum_{i=1}^n\frac{1}{1+\sum_{j=1,j\ne i}^n1}-\sum_{i=1}^n\frac{1}{(n-1)+1}\\=n\times\frac1{1+(n-1)}-n\times\frac1{(n-1)+1}=0$$ Which again is ...

Partial Proof: For general case of n variables, the inequality converts to:

$$\sum_i^n \frac1{1+a_1+a_2+\cdots+a_n-a_i}\le \sum_i^n\frac1{n-1+a_i}$$

Similiar to the given proof we can convert $\frac {a_1}{a_1+(n-1)}$ like this: $$\frac{a_1}{a_1+(n-1)}=\frac{a_1}{a_1+(n-1)(a_1a_2\cdots a_n)^{1/n}}$$ Dividing numerator and denominator by $a_1^{1/n}$: $$=\frac{a_1^{(n-1)/n}}{a_1^{(n-1)/n}+(n-1)\left(\frac{a_1a_2\cdots a_n}{a_1}\right)^{1/n}}$$ Finally using AM-GM gives: $$\frac{a_2+a_3+\cdots+a_n}{(n-1)}\ge\left(a_2a_3\cdots a_n\right)^{1/(n-1)}$$ Or: $$(n-1)\left(a_2a_3\cdots a_n\right)\le (a_2+a_3+\cdots+a_n)^{n-1}$$ $$=\frac{a_1^{(n-1)/n}}{a_1^{(n-1)/n}+(n-1)\left(\frac{a_1a_2\cdots a_n}{a_1}\right)^{1/n}} \ge\frac{a_1^{(n-1)/n}}{a_1^{(n-1)/n}+a_2^{(n-1)/n}+\cdots+a_n^{(n-1)/n}}$$

So, $$\sum_i^n\frac{a_i}{a_i+(n-1)}\ge\sum_i^n\frac{a_1^{(n-1)/n}}{a_1^{(n-1)/n}+a_2^{(n-1)/n}+\cdots+a_n^{(n-1)/n}}=1\tag{i}$$ We have proved what we need for the general case of n-variables, try putting $n=3$.

Since product of all numbers is 1, we can define new fractions as: Let $$\displaystyle a_1:=\frac{x_1}{x_2},a_2:=\frac{x_2}{x_3},\cdots,a_n:=\frac{x_n}{x_1}$$

Notice that in the given proof: $$\frac{b}{ab+b+1}=\frac{x_2/x_3}{x_1/x_2.x_2/x_3+x_2/x_3+1}=\frac{x_2}{x_1+x_2+x_3 }$$

Now similiar to the given proof we can show that(step unproven): $$\frac{2}{a_1+(n-1)}-\frac{1}{1+a_2+a_3+\cdots+a_n}-\frac{x_2}{x_!+x_2+\cdots+x_n }\ge0$$ $$\frac{2}{a_1+(n-1)}\ge \frac{1}{1+a_2+a_3+\cdots+a_n}+\frac{x_2}{x_1+x_2+\cdots+x_n }$$

$$\frac1{a_1+(n-1)}+\frac1{a_1+(n-1)}\ge\frac{1}{1+a_2+a_3+\cdots+a_n}+\frac{x_2}{x_1+x_2+\cdots+x_n }$$ Since $\displaystyle \sum_i^n\frac{x_2}{x_1+x_2+\cdots+x_n }=1$ $$\frac1{a_1+(n-1)}+\frac1{a_1+(n-1)}\ge\frac{1}{1+a_2+a_3+\cdots+a_n}+1$$ $$\sum_{cyc}\frac{1}{a_1+(n-1)}-\sum_{cyc,i}\frac1{1+a_2+a_3+\cdots+a_n }\ge1-\sum_{cyc}\frac{1}{a_1+(n-1)}\ge0$$ The$\ge0$ part, we have proved in (i). So, $${\large \sum_{cyc}\frac{1}{a_1+(n-1)}\ge\sum_{cyc,i}\frac1{1+\sum_{cyc,j\ne i}a_j}}\Box$$

• why the downvote? – RE60K Aug 28 '14 at 6:50
• Complaint sustained. Two downvotes and nobody explaining what's wrong. Why? – Han de Bruijn Aug 30 '14 at 10:31
• @math110 added more details – RE60K Aug 30 '14 at 12:02
• I haven't downvoted this yet, but if I were to hazard a guess, I would say that the "proof" is sketchy and going back-and-forth a lot. You could improve it by having a clean logical flow from start to finish. Also you need to prove statements rather than simply stating they can be extrapolated. For e.g. there is no proof I observe for a key statement used: $$\frac{2}{a_1+(n-1)}-\frac{1}{1+a_2+a_3+\cdots+a_n}-\frac{x_2}{x_!+x_2+\cdots+x_n }\ge 0$$ which to me at least is not obvious. – Macavity Aug 30 '14 at 14:02
• @Macavity Hmm I did'nt thought about that example – RE60K Aug 30 '14 at 14:16