A strengthening of a known inequality; looking for a neater solution

Question

If $a\ge b\ge c > 0$ then

$$ \begin{split} \frac{(a-b)^2}{(a+b)} + \frac{(b-c)^2}{(b+c)} & \ge \sqrt{3(a^2+b^2+c^2)}- (a+b+c) \\&\ge \frac{(a-b)^2}{\frac{1+\sqrt{3}}{2}a+\frac{5-\sqrt{3}}{2} b} +\frac{(b-c)^2}{(1+\sqrt{\frac{3}{2}})b+(2-\sqrt{\frac{3}{2}}) c} \end{split} $$

This is a strengthening of an inequality found here.

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Notes:

1. Found the RHS inequality by choosing the coefficients in such a way that (after a substitution $a=u+v+w$, $b=u+v$, $c=u$), the coefficients of all of the monomials in $u$, $v$, $w$ in the numerator of the difference of the squares of sides are $\ge 0$ ( also known as the Buffalo way, but it goes back at least to Adolf Hurwitz). The possible coefficients in the denominator form a convex set of a fairly simple kind ( a product in fact). The choices are those of a vertex. One can do that for every $n$, the formulas are not that complicated.

2. Found the LHS inequality by numerical optimization. In fact, one cannot do it with the previous method.

3. A general comment: one could compose many inequalities about the extreme of some algebraic function on some semialgebraic set. In a fixed dimension the answer can be given in principle as a real algebraic number ( if not infinite). For inequalities that involve $n$ unknowns, where $n$ is arbitrary, things can be complicated. Some , ( the mean inequalities, or the Newton/Maclaurin inequalties $\&$c) are established for all $n$. It may not be feasible to do that in general.

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At this stage I don't see a neater way to prove it. Any feedback would be appreciated!

In a previous version the LHS was not correct, I thank @River Li: for pointing out that.

Answer 1

Partial answer.

• For the left inequality

It suffices to prove that $$\left(\frac{(a-b)^2}{(a+b)} + \frac{(b-c)^2}{(b+c)} + (a + b + c)\right)^2\ge 3(a^2 + b^2 + c^2)$$ or \begin{align*} &\left(\frac{(a-b)^2}{(a+b)} + \frac{(b-c)^2}{(b+c)}\right)^2 + 2\left(\frac{(a-b)^2}{(a+b)} + \frac{(b-c)^2}{(b+c)}\right)(a + b + c)\\[6pt] &\qquad + (a + b + c)^2 \ge 3(a^2 + b^2 + c^2) \end{align*} or (using $(u+v)^2 \ge 4uv$) $$\begin{align*} &4\cdot \frac{(a-b)^2}{(a+b)} \cdot \frac{(b-c)^2}{(b+c)} + 2\left(\frac{(a-b)^2}{(a+b)} + \frac{(b-c)^2}{(b+c)}\right)(a + b + c)\\[6pt] &\qquad + (a + b + c)^2 \ge 3(a^2 + b^2 + c^2) \end{align*}$$ or (multiplying both sides by $(a+b)(b+c)$) $$ \left( 5\,{a}^{2}-5\,ab+2\,{b}^{2} \right) {c}^{2}+ \left( -5\,{a}^{2 }b+4\,a{b}^{2}-3\,{b}^{3} \right) c+2\,{a}^{2}{b}^{2}-3\,a{b}^{3}+3\,{ b}^{4} \ge 0$$ which is true since $5\,{a}^{2}-5\,ab+2\,{b}^{2} \ge 0$ and \begin{align*} &4\cdot \left( 5\,{a}^{2}-5\,ab+2\,{b}^{2} \right) \cdot (2\,{a}^{2}{b}^{2}-3\,a{b}^{3}+3\,{ b}^{4})\\ &\qquad - (-5\,{a}^{2 }b+4\,a{b}^{2}-3\,{b}^{3})^2\\ ={}& 15(a-b)^4b^2 \ge 0. \end{align*}

We are done.