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I've been playing around with Nesbitt's inequality and I can see that it must be true.

Nesbitt's inequality states that for positive real numbers $a$, $b$, and $c$:

$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq \frac{3}{2}$$

I'm a leyman and saw some proofs using the Cauchy-Schwarz inequality and the Jensen's inequality but I don't understand those. Is there any easier way?

Jose Avilez
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Bro Biden
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  • The short answer is yes. I saw such a proof in the first few pages the book “Inequalities” by Cvetkovski. The proof isn’t super obvious though. – Adam Rubinson Jan 25 '22 at 05:36
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    There are nine different proofs in https://en.wikipedia.org/wiki/Nesbitt%27s_inequality – Martin R Jan 25 '22 at 05:37
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    See the third proof in the Wikipedia page. Hint: if your question has an answer that is easy to find googling or in Wikipedia, expect to receive downvotes. – jjagmath Jan 25 '22 at 05:39
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    It would be more constructive if you tell us what proofs you have seen and what you do not understand about them. Otherwise there is a high risk that another proof presented here will also not satisfy you. – Martin R Jan 25 '22 at 05:40

1 Answers1

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There are many way to prove Nesbitt's ineqality. I hope this one works for you.

From Nesbitt's inequality it directly follows that $$\frac{a^3 + a^2 b + a^2 c + a b^2 + 3 a b c + a c^2 + b^3 + b^2 c + b c^2 + c^3}{(a + b) (a + c) (b + c)} \geq \frac{3}{2}$$ And by clearing denominators we get $$2(a^3 + a^2 b + a^2 c + a b^2 + 3 a b c + a c^2 + b^3 + b^2 c + b c^2 + c^3) \\ \geq 3(a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2)$$ This holds if and only if $$2 a^3 + 2 b^3 + 2 c^3 \geq a^2 b + a^2 c + a b^2 + a c^2 + b^2 c + b c^2\text{ }\text{ }\text{ }\text{ }(\star)$$ By Cauchy's Inequality (AM-GM) we have $$\begin{equation} \frac{a^3 + a^3 + b^3}{3} \geq \sqrt[3]{a^3 a^3 b^3} = a^2 b \end{equation}$$ Using this it follows $$\begin{align*} 2 a^3 &+ 2 b^3 + 2 c^3 \\ &= \frac{a^3 + a^3 + b^3}{3} + \frac{a^3 + a^3 + c^3}{3} + \frac{a^3 + b^3 + b^3}{3} \\ &+ \frac{a^3 + c^3 + c^3}{3} + \frac{b^3 + b^3 + c^3}{3} + \frac{b^3 + c^3 + c^3}{3} \\ &\geq a^2 b + a^2 c + a b^2 + a c^2 + b^2 c + b c^2 \end{align*}$$ This proves $(\star)$ and thus the desired Nesbitt's inequality.

Yanick
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