# Proof of Nesbitt's Inequality: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$?

I just thought of this proof but I can't seem to get it to work.

Let $a,b,c>0$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$

Proof: Since the inequality is homogeneous, WLOG $a+b+c=1$. So $b+c=1-a$, and similarly for $b,c$. Hence it suffices to prove that $$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\ge \frac{3}{2}$$. From $a+b+c=1$ and $a,b,c>0$ we have $0<a,b,c<1$, so we have $$\frac{a}{1-a}=a+a^2+a^3+...$$ and similarly for $b,c$, so it suffices to prove that $$\sum_{cyc} a+\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{3}{2}$$, or equivalently (by $a+b+c=1$) $$\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{1}{2}$$, where $\sum_{cyc} a=a+b+c$ similarly for $\sum_{cyc}a^n=a^n+b^n+c^n$. Here I get stuck. For example, doing the stuff below yields a weak inequality, because of too many applications of the $a^2+b^2+c^2\ge (a+b+c)^2/3$ inequality.

"stuff below": Now, from $0<a<1$ we have $a^3>a^4$, $a^5>a^8$, $a^7>a^8$, $a^9>a^{16}$, and so on, so it suffices to prove that $$\sum_{cyc} a^2+2\sum_{cyc} a^4+4\sum_{cyc} a^8+8\sum_{cyc} a^{16}+...\ge \frac{1}{2}$$, or, multiplying by 2, $$2\sum_{cyc} a^2+4\sum_{cyc} a^4+8\sum_{cyc} a^8+...\ge 1$$, which by a simple inequality (i.e. recursively using $a^{2^n}+b^{2^n}+c^{2^n}\ge \frac{(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2}{3}$) is equivalent to $$2(1/3)+4(1/3)^3+8(1/3)^7+...+2^n(1/3)^{2^n-1}+...\ge 1$$. But then http://www.wolframalpha.com/input/?i=%5Csum_%7Bi%3D1%7D%5E%7B5859879%7D+2%5Ei+*%281%2F3%29%5E%282%5Ei-1%29 and so we're screwed.

• Just in case you did not notice, this inequality can be proven by concave property of f(x)=\frac{x}{1-x} or making left hand side of the original inequality common denominator(then cross multiply to get the equivalent polynomial inequality). – user175968 Oct 19 '14 at 12:00
• – Macavity Oct 19 '14 at 12:25
• @AnalysisIncarnate Is it okay to prove it using AM-GM Inequality? If so see my answer – user171358 Oct 19 '14 at 13:12
• Something seems to be wrong with the Wolfram Alpha link. – Nate Eldredge Oct 19 '14 at 14:52
• – Martin Sleziak Aug 10 '17 at 6:47

If you want to use the power series way to do it(although I think it's overkill), by convex property of $f(x)=x^n$ we have $\frac{(a^n+b^n+c^n)}{3}\geq (\frac{a+b+c}{3})^n$ or equivalently $(a^n+b^n+c^n)\geq \frac{(a+b+c)^n}{3^{n-1}}$, where $n$ is positive integer.

Therefore your left hand side(series expansion) is greater or equal to $1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}...=\frac{3}{2}$

I'm sorry but AM-GM Inequality is all that I know, So, By using AM-GM Inequality for $(a+b) , (b+c) , (c+a)$ and $\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$ We get $$\frac{(a+b)+(b+c)+(c+a)}{3} \ge\sqrt[3]{(a+b)(b+c)(c+a)}\tag{1}$$ $$\frac{\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}}{3} \ge\sqrt[3]{\frac{1}{(b+c)(c+a)(a+b)}}\tag{2}$$ Multiplying $(1)$ and $(2)$ We get. $$\frac{1}{9}\left[(a+b)+(b+c)+(c+a)\right]\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{(b+c)(c+a)(a+b)}}$$ $$\left[(a+b)+(b+c)+(c+a)\right]\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge9$$ After Expanding $$3+\frac{a+b}{b+c}+\frac{c+a}{b+c}+ \frac{a+b}{c+a}+\frac{b+c}{c+a}+ \frac{b+c}{a+b}+\frac{c+a}{a+b} \ge\ 9$$ By rearrangement $$\left(\frac{a+b}{b+c}+\frac{c+a}{b+c}\right)+ \left(\frac{a+b}{c+a}+\frac{b+c}{c+a}\right)+ \left(\frac{b+c}{a+b}+\frac{c+a}{a+b}\right) \ge\ 6$$ $$\frac{2a+b+c}{b+c}+ \frac{a+2b+c}{c+a}+ \frac{b+2c+a}{a+b} \ge\ 6$$ $$\frac{2a}{b+c}+\frac{b+c}{b+c}+ \frac{2b}{c+a}+\frac{a+c}{c+a}+ \frac{2c}{a+b}+\frac{b+a}{a+b} \ge\ 6$$ $$3+\frac{2a}{b+c}+ \frac{2b}{c+a}+ \frac{2c}{a+b} \ge\ 6$$ $$\frac{2a}{b+c}+ \frac{2b}{c+a}+ \frac{2c}{a+b} \ge\ 3$$ $$\frac{a}{b+c}+ \frac{b}{c+a}+ \frac{c}{a+b} \ge\ \frac{3}{2}$$ As Simple as that

First Proof

$a+b=x$ , $b+c=y$ , $c+a=z$

$\therefore 2 \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=(x+y+z)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-6\\=\underbrace{\dfrac{x}{y}+\dfrac{y}{x}}_{\geq 2}+\underbrace{\dfrac{y}{z}+\dfrac{z}{y}}_{\geq 2}+\underbrace{\dfrac{z}{x}+\dfrac{x}{z}}_{\geq 2}-3 \geq 3$

Second Proof

$f(a,b,c)=\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right) \implies f(ta,tb,tc)=f(a,b,c)$ $\forall t\in \mathbb{R}\setminus\{0\}$ $a+b+c=1 \implies f(a,b,c)=\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3 \geq \dfrac{9}{2}-3=\dfrac{3}{2}$

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

Now M + N = 3.

M + S >= 3 (AM-GM Property)

N + S >= 3 (AM-GM Property)

M + N + 2S >= 6 (Adding above inequalities)

2S >= 3

S >= $\frac{3}{2}$

$$\sum_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b-(c-a)}{2(b+c)}=$$ $$=\sum_{cyc}(a-b)\left(\frac{1}{2(b+c)}-\frac{1}{2(a+c)}\right)=\sum_{cyc}\frac{(a-b)^2}{2(a+c)(b+c)}\geq0.$$ Done!

Observe that

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2} \iff (\frac{a}{b+c}+1)+(\frac{b}{c+a}+1)+(\frac{c}{a+b}+1)= (a+b+c)(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b})\ge \frac{9}{2}.$

By Titu's lemma

$\frac{1^2}{b+c}+\frac{1^2}{c+a}+\frac{1^2}{a+b}\ge \frac{(1+1+1)^2}{(a+b)+(b+c)+(c+a)} = \frac{9}{2(a+b+c)}$.

So

$(a+b+c)(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b})\ge (a+b+c)\frac{9}{2(a+b+c)} = \frac{9}{2}.$