# Nesbitt's Inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$

I'm reading a book which focus in inequality.

I'm stuck in this question. Let $a,b,c$ be positive real numbers. (Nesbitt's inequality) Prove the inequality $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$$

So the first step of solution given is $\frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{a+c}{c+b}+\frac{c+b}{a+c}+\frac{b+a}{a+c}+\frac{a+c}{b+a}\geq2+2+2=6$

I don't know how to proceed from the question to the first step of solution. Can anyone explain?

• The last term on the LHS of the inequality should be $\frac{a+c}{b+a}$ – J.Gudal Dec 26 '15 at 17:58
• Possible duplicate – A.Γ. Dec 26 '15 at 19:09

The first step of the solution says that $X+\frac1X+Y+\frac1Y+Z+\frac1Z\geq2+2+2$.
where $X=\frac{a+b}{b+c}$ and so on.
Do you know that $X+\frac1X$ is always $2$ or more whenever $X>0$?

• I've understood. Thanks – Mathxx Dec 26 '15 at 17:58
• Can I ask why $X+\frac{1}{X}$ is always greater and equal to 2? – Mathxx Dec 26 '15 at 18:00
• If you believe the AM-GM inequality, then it follows from the fact that the geometric mean of $X$ and $1/X$ is $1$, so the arithmetic mean is at least $1$. – Ben Millwood Dec 26 '15 at 18:07
• @Mathxx if $\; X \ge 1\;$ then $\frac{1}{X}+X =\frac{X^2+X}{X} \ge \frac{X+X}{X} \ge 2\;$ If $X \le 1\;$then $X\;$can be written as $X=\frac{1}{Y}\;$ where $\; Y \ge 1 \;$. Hence, $\; \frac{1}{X}+X=\frac{1}{Y}+Y \ge 2$ – XPenguen Dec 26 '15 at 18:21

HINT: set $$b+c=x$$ $$c+a=y$$ $$a+b=z$$ adding we get $$a+b+c=\frac{x+y+z}{2}$$ and we can compute $$a+x=\frac{x+y+z}{2}$$ thus $$a=\frac{-x+y+z}{2}$$ etc

We have three expressions: $$S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$$

$$M=\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}$$

$$N=\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}$$ Obviously, we have $M+N=3$ According to AM-GM,we have:

$$M+S=\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}\geqslant 3$$ $$N+S=\frac{a+c}{b+c}+\frac{a+b}{c+a}+\frac{b+c}{a+b}\geqslant 3$$ Thus, $M+N+2S\geqslant 6$ so $2S\geqslant 3$ so $S\geqslant \dfrac{3}{2}$ q.e.d

More easier and obvious answer: $$\frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+c}\geq\frac{3}{2}$$ $$\frac{2a}{b+c}+\frac{2c}{a+b}+\frac{2b}{a+c}\geq3$$ $$\frac{2a+b+c}{b+c}+\frac{b+2c+a}{a+b}+\frac{c+2b+a}{a+c}\geq6$$ $$\frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{a+c}{c+b}+\frac{c+b}{a+c}+\frac{b+a}{a+c}+\frac{a+c}{b+a}\geq6$$

• I don't think this addresses the questioner's specific problem at all. – Ben Millwood Dec 26 '15 at 18:11
• I started from the first step that he gave – Olimjon Dec 26 '15 at 18:12
• Yes. The questioner wants to know how to get TO the first step. – Ben Millwood Dec 26 '15 at 18:12
• In this case, just reversing is right. – Olimjon Dec 26 '15 at 18:13