Stronger than Nesbitt's inequality using convexity and functions

Question

Hi it's a refinement of Nesbitt's inequality and for that, we introduce the function :

$$f(x)=\frac{x}{a+b}+\frac{b}{x+a}+\frac{a}{b+x}$$

With $a,b,x>0$

Due to homogeneity we assume $a+b=1$ and we introduce the function :

$$g(a)=\frac{a}{1-a+x}$$

Showing that $g(a)$ is convex on $(0,1)$ is not hard so we have :

$$\frac{b}{x+a}+\frac{a}{b+x}\geq 2\frac{a+b}{2(\frac{a+b}{2}+x)}=h(x)$$

So we have :

$$f(x)\geq h(x)+\frac{x}{a+b}$$

Now we put $u=\frac{x}{a+b}$ and we want to show :

$$h(x)+\frac{x}{a+b}=u+\frac{1}{0.5+u}\geq \frac{3}{2}$$

The last inequality is obvious.

My question :

It is correct?

Do you know other refinements?

Thanks in advance!

Ps: I add the tag reference request for the last question.

Michael Rozenberg · Accepted Answer · 2020-11-01T17:07:01.170

1

There are very many refinements of the Nesbitt's inequality.

For example. For positives $a$, $b$ and $c$ we have:

1.$$\frac{a}{\sqrt[3]{4(b^3+c^3)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2};$$ 2.$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{(a+b+c)^2}{2(ab+ac+bc)};$$ 3.$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3\sqrt{5(a^2+b^2+c^2)-ab-ac-bc}}{4(a+b+c)};$$ 4. For any reals $a$, $b$ and $c$ such that $ab+ac+bc>0$ prove that: $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}.$$

edited Nov 01 '20 at 17:07

answered Nov 01 '20 at 16:39

Michael Rozenberg

194,933

$u+\frac{1}{0.5+u}= u+.5+\frac{1}{0.5+u}-.5 \geq 2-.5$? – Neat Math Nov 01 '20 at 16:48
@Neat Math I did not understand. What is your question? – Michael Rozenberg Nov 01 '20 at 16:57
I'm saying your calculation $(u+1)(u-2)\ge 0$ is not right because if you plug in 2 or -1 in the last inequality it doesn't become equality. I'm showing OP's right on that one. It's equivalent to $(2u-1)^2 \geq 0$. – Neat Math Nov 01 '20 at 17:01
@Neat Math It was my delirium. Thank you! – Michael Rozenberg Nov 01 '20 at 17:09
@MichaelRozenberg Thanks you ! I really like the 4. – Miss and Mister cassoulet char Nov 01 '20 at 17:22
@Erik Satie You are welcome! The first inequality was in the Israeli test. – Michael Rozenberg Nov 01 '20 at 17:26
@MichaelRozenberg I think there is a typo in your #3. When $a=b=c$ the RHS is not 3/2. – Neat Math Nov 02 '20 at 00:31
also this: https://math.stackexchange.com/questions/2281703/stronger-than-nesbitts-inequality-fraca-sqrt48b4c4-fracbac – River Li Nov 02 '20 at 02:31

score 1 · Answer 2 · answered Nov 02 '20 at 06:21

Stronger than Nesbitts inequality for non-cyclic style

If $a,\,b,\,c$ are positive real numbers, then $$ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{a+b}{b+c}+\frac{b+c}{a+b}+1$$
If $a,\,b,\,c$ are positive real numbers, then $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}+\frac{27(b-c)^{2}}{16(a+b+c)^{2}}.$$
Let $a,\,b,\,c$ be nonnegative real numbers, no two of which are zero. Prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}+\frac{(b-c)^{2}}{2(b+c)^{2}}.$$
Let $a,b,c$ are non-negative real numbers such that $ab+bc+ca>0.$ Prove that $$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} \geqslant \dfrac{3}{2}+\dfrac{7(b-c)^2}{16a(b+c)+7bc}.$$

Stronger than Nesbitt's inequality using convexity and functions

2 Answers2