Conjecture about :$\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac94+\frac32 \cdot \frac{(a-b)^2}{ab+bc+ca}}$

Question

It's a conjecture . I combine two inequalities : Stronger than Nesbitt inequality and Stronger than Nesbitt's inequality $\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}$

So now the conjecture :

Let $a,b,c>0$ such that $a\geq b \geq c$ then we have :

$$\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac94+\frac32 \cdot \frac{(a-b)^2}{ab+bc+ca}}$$

I cannot find any counter-example until now . To prove it I have tried the same method as here Prove this refinement of Nesbitt's inequality based on another without success .

My question :

Have you an element of proof or an counter-example ?

Thanks in advance !!

Max .

@RiverLi have you a counter-example or a proof ? Thanks!! – Miss and Mister cassoulet char Nov 14 '20 at 17:15 — Miss and Mister cassoulet char, Nov 14 '20 at 17:15
Maybe Michael Rozenberg's shadow have downvoted ... – Miss and Mister cassoulet char Nov 14 '20 at 17:32 — Miss and Mister cassoulet char, Nov 14 '20 at 17:32

score 4 · Accepted Answer · edited Jan 20 '21 at 00:38

4

There is a good reason that you can not prove: $$\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac94+\frac32 \cdot \frac{(a-b)^2}{ab+bc+ca}}$$ It's wrong for $a=\frac53,b=1,c=\frac19.$ By the way, the following inequality is true: $$\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac94+\frac{(a-b)^2}{ab+bc+ca}},$$ as verified by Bottema - Maple program.

edited Jan 20 '21 at 00:38

Hans

9,804

answered Nov 20 '20 at 07:37

NKellira

2,001

2

Powerful Bottema! – River Li Nov 21 '20 at 02:09

Conjecture about :$\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac94+\frac32 \cdot \frac{(a-b)^2}{ab+bc+ca}}$

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