# Prove of Nesbitt's inequality in 6 variables

I was just reading "Pham kim hung secrets in inequalities,Volume 1" book and there was an interesting problem on it's Cauchy-Schwarz and Holder section that caught my eye.

Prove that for all positive real numbers $a,b,c,d,e,f$,we always have $$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d}{e+f}+\frac{e}{f+a}+\frac{f}{a+b}\ge 3$$

The writer of the book proved it by Cauchy-Schwarz,But there was an another method in the start of the book for proving original Nesbitt inequality.

Prove that for all non-negative real numbers $a,b,c$ $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}$$ Solution: set $S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$,$M=\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}$,$N=\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}$. obliviously $M+N=3$.and by AM-GM we get the $$M+S\ge3,N+S\ge3$$ So $M+N+2S\ge6$ and $M+N=3$ we get $S\ge\frac{3}{2}$.

Like this method (calling it $S,M,N$),He proves $$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2$$

for all non-negative $a,b,c,d$.

As I Liked this method, I started to proof the 6 variable variation using it. $$S=\sum\limits_{cyc}\frac{a}{b+c}$$ $$M=\sum\limits_{cyc}\frac{b}{b+c}$$ $$N=\sum\limits_{cyc}\frac{c}{b+c}$$

it easy too see that $M+N=6$.and using AM-GM it is easy to reach that $M+S\ge6$ but for proving $N+S \ge 6$ $$N+S=\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+e}{d+e}+\frac{d+f}{e+f}+\frac{e+a}{f+a}+\frac{f+b}{a+b}\ge 6$$

I don't see any way to change this to something easy to work with it.By the way it is strange that why it is called Nesbitt's 6 variables inequality in book because Nesbitt Generalization is $$\sum_{i=1}^{n}\frac{a_i}{s-a_i}\ge\frac{n}{n-1}$$ Where $\sum_{i=1}^{n}a_i = s$ for positive $a_1,\ldots a_n$.

• There are many ways to generalise Nesbitt's inequality. This way is perhaps more aptly called Shapiro's inequality (en.wikipedia.org/wiki/Shapiro_inequality). As you may see from the link, for larger $n$, the inequality may not even hold. It does for $n=6$. – Macavity Sep 7 '14 at 3:13

$$\sum\limits_{i=1}^{6}\frac{x_i}{x_{i+1}+x_{i+2}} \ge 3$$ where, the indices are taken cyclically. Wlog, assume $\sum\limits_{i=1}^{6} x_i = 1$.

We start with the fact that $f(s) = \frac{1}{1-s}$ is convex on the interval $[0,1)$. Applying Jensen Inequality, $$\sum\limits_{i=1}^{6} \frac{x_i}{1-(x_i + x_{i-1}+x_{i-2}+x_{i-3})} \ge \frac{1}{1-\sum\limits_{i=1}^{6}x_i(x_i + x_{i-1}+x_{i-2}+x_{i-3})}$$

$$\iff \sum\limits_{i=1}^{6}\frac{x_i}{x_{i+1}+x_{i+2}} \ge \frac{1}{\left(\sum\limits_{i=1}^6 x_i\right)^2-\sum\limits_{i=1}^{6}x_i(x_i + x_{i-1}+x_{i-2}+x_{i-3})}$$

Now, $\displaystyle \left(\sum\limits_{i=1}^6 x_i\right)^2-\sum\limits_{i=1}^{6}x_i(x_i + x_{i-1}+x_{i-2}+x_{i-3}) = (x_1+x_4)(x_3+x_6)+(x_1+x_4)(x_2+x_5)+(x_2+x_5)(x_3+x_6)$

Since, $\displaystyle 1 = \left((x_1+x_4)+(x_3+x_6)+(x_2+x_5)\right)^2 \ge 3\left((x_1+x_4)(x_3+x_6)+(x_1+x_4)(x_2+x_5)+(x_2+x_5)(x_3+x_6)\right)$

the desired inequality follows.

As for the $N+S \ge 6$ inequality, is not always true.

Take the values $(a,b,c,d,e,f) = (1,0,1.02,0.02,1.01,0.02)$.

Then, $N+S = 5.99991989925 < 6$ (when I say $b = 0$ I mean to take a positive value as close to $0$ as possible)

• This is a proof of Shapiro's for six variables. However it leaves open whether or not the inequality $N+S \ge 6,$ in the notation of the posted question. [I haven't been able to deduce the latter, even if Shapiro's is assumed...] – coffeemath Sep 8 '14 at 13:00
• @coffeemath $N+S \ge 6$ may be correct! but the Quadratic form obtained after applying Cauchy-Schwarz doesn't look positive(semi-definite) (not all characteristic roots are positive). Where as QF for Shapiro-6 is clearly a positive(semidefinite) form. Seems its stronger than Shapiro-6 Ineq :o – r9m Sep 13 '14 at 0:23
• I'm probably being dense, but how does one set up Cauchy-Schwartz ["CS"] for inequalities of this type? For CS one gets a dot product bounded above, rather than below, and I couldn't see the connection. – coffeemath Sep 14 '14 at 6:31
• @coffeemath $\displaystyle \left(\sum\limits_{i=1}^{6}\frac{x_i}{x_{i+1}+x_{i+2}}\right)\left(\sum\limits_{i=1}^{6} x_i(x_{i+1}+x_{i+2})\right) \ge \left(\sum\limits_{i=1}^{6} x_i\right)^2$ by CS. I have tried to elaborate why CS fails for these here. – r9m Sep 14 '14 at 7:48
• @r9m,thank you for your answer and comments but i get confused a little about the part you say $N+S \ge 6$ maybe stronger than Shapiro 6 inequality.would you please add explanation about this to your answer? – user2838619 Sep 14 '14 at 8:41