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Suppose $a>0, b>0, c>0$.

Prove that: $$a+b+c \ge \frac{3}{2}\cdot [(a+b)(a+c)(b+c)]^{\frac{1}{3}}$$

Hence or otherwise prove: $$\color{blue}{\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}}$$

Dr. Mathva
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M.You
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  • The final inequality you want to prove is known as Nesbitt's inequality. See the answers here or here for proofs. – Minus One-Twelfth Mar 30 '19 at 11:50
  • Also, the first inequality you are asked to prove follows from the AM-GM inequality ($\frac{x+y+z}{3}\ge \sqrt[3]{xyz}$ for $x,y,z\ge 0$). – Minus One-Twelfth Mar 30 '19 at 11:56
  • @MinusOne-Twelfth Does this name Nesbitt come from Perso-Arabic نسبت (Transliterated as nesbat or nisbat) which means relation? –  Oct 07 '20 at 17:00

4 Answers4

4

Hint: Substitute $$b+c=x,a+c=y,a+b=z$$ so $$a=\frac{-x+y+z}{2}$$ $$b=\frac{x-y+z}{2}$$ $$c=\frac{x+y-z}{2}$$ And we get $$\frac{-x+y+z}{2x}+\frac{x-y+z}{2y}+\frac{x+y-z}{2z}\geq \frac{3}{2}$$ Can you finish? And we get $$\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{x}{z}+\frac{z}{x}\geq 6$$

Dr. Sonnhard Graubner
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3

Using AM-GM inequality:$$\frac{x_1+\dots+x_n}{n}\geq (x_1\dots x_n)^{1/n}$$ let $n=3$ and $x_1=a+b,\; x_2=b+c,\; x_3=c+a$: $$\frac{2(a+b+c)}{3}\geq [(a+b)(b+c)(c+a)]^{1/3}$$

user376343
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Xin Fu
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3

Hint:

$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}\iff \frac{a+b+c}{b+c}+\frac{b+a+c}{a+c}+\frac{c+a+b}{a+b}\ge \frac{3}{2}+3$$

$$\iff \big(a+b+c\big)\cdot \bigg(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\bigg)\ge \frac{9}{2}\iff \color{blue}{\frac{2\cdot (a+b+c)}{3}\ge \frac{3}{\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}}}$$ Which is trivial by the AM-HM inequality. Done!

Dr. Mathva
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2

To minimize $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ , we need to have $$ \begin{align} 0 &=\delta\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\\ &=\left(\frac1{b+c}-\frac{b}{(c+a)^2}-\frac{c}{(a+b)^2}\right)\delta a\\ &+\left(\frac1{c+a}-\frac{c}{(a+b)^2}-\frac{a}{(b+c)^2}\right)\delta b\\ &+\left(\frac1{a+b}-\frac{a}{(b+c)^2}-\frac{b}{(c+a)^2}\right)\delta c\tag1 \end{align} $$ for all $\delta a,\delta b,\delta c$. That means $$ \begin{align} \frac1{a+b}&=\frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}\tag2\\ \frac1{b+c}&=\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}\tag3\\ \frac1{c+a}&=\frac{c}{(a+b)^2}+\frac{a}{(b+c)^2}\tag4 \end{align} $$ Subtract $(4)$ from the sum of $(2)$ and $(3)$: $$ \frac1{b+c}+\frac1{a+b}-\frac1{c+a}=\frac{2b}{(c+a)^2}\tag5 $$ Add $\frac2{c+a}$ and divide by $2(a+b+c)$: $$ \frac1{2(a+b+c)}\left(\frac1{b+c}+\frac1{a+b}+\frac1{c+a}\right)=\frac1{(c+a)^2}\tag6 $$ By symmetry, $$ \frac1{(a+b)^2}=\frac1{(b+c)^2}=\frac1{(c+a)^2}\tag7 $$ from which we get $a=b=c$. Thus, we get $$ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac32\tag8 $$

robjohn
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