Proof of a tighter inequality than Cauchy-Schwarz inequality

Question

A few days ago, I found this post which discuss about some tighter versions of Cauchy-Schwarz (or some might prefer the name AM-GM) inequality. User @Michael Rozenberg proposed a very interesting inequality at the comment section

Let $a$, $b$ and $c$ be three positive numbers, then the following inequality holds $$ 3(a^2+b^2+c^2) - (a+b+c)^2 \ge \dfrac{25(a-b)^2(b-c)^2(c-a)^2}{a^4+b^4+c^4}. $$

He also mentioned that if we replace number $25$ at the R.H.S with $26$ then we'll get a wrong inequality, so perhaps $25$ is the "best" number, although I don't know what happen if we replace with $25 + \varepsilon$ for $\varepsilon > 0$ sufficiently small.

What I'm curious about is how to fully prove the above inequality. What I tried is to use $pqr$ method: Let $p = a + b + c$, $q = ab + bc + ca$, $r = abc$ and without loss of generality, we can assume that $p = 1$: Since if $p = M$ for some real number $M$, using the change of variables $a' = \frac{a}{M}$, $b' = \frac{b}{M}$ and $c' = \frac{c}{M}$, we get the same inequality as before but this time $a' + b' + c' = 1$. With the $pqr$ substitution and the assumption $p = 1$, we'll have \begin{align*} a^2 + b^2 + c^2 &= 1 - 2q \\ a^4 + b^4 + c^4 &= 2q^2 - 4q + 4r + 1 \\ (a-b)^2(b-c)^2(c-a)^2 &= -4q^3 + q^2 + 18qr - 27r^2 - 4r \end{align*} The inequality we need to prove now is $$ 104q^3 - 13q^2 - 458qr + 675r^2 - 10q + 108r + 2 \ge 0. $$ My idea to continue from here is to use Schur inequality of order 1: $r \ge \max\{0, \frac{1}{9}(4q-1)\}$, but this idea is not working since the L.H.S contains the term $-458qr$ makes "$\ge$" sign become "$\leq$" sign. I still don't know how to continue from here, and I'm not sure if $pqr$ method is a good idea.

There are three questions that I really want the answer:

(1) How can we prove the above inequality?

(2) How do we know that $25$ is the "best" constant for the R.H.S?

(3) Is there any other function $f(a,b,c) \ge 0$ such that $$3(a^2 + b^2 + c^2) - (a+b+c)^2 \ge f(a,b,c), \quad \forall a,b,c > 0?$$ If there exists, how do we actually come up with such function?

Answer 1

1. Let $c=\min\{a,b,c\}$,$a=c+u$,$b=c+v$ and $u^2+v^2=2tuv$.

Thus, $$(a^4+b^4+c^4)(3(a^2+b^2+c^2)-(a+b+c)^2)-25\prod_{cyc}(a-b)^2=$$ $$=((c+u)^4+(c+v)^4+c^4)\sum_{cyc}(a-b)^2-25\prod_{cyc}(a-b)^2\geq$$ $$=(u^4+v^4)(u^2+v^2+(u-v)^2)-25(u-v)^2u^2v^2=$$ $$=u^3v^3((4t^2-2)(4t-2)-25(2t-2))=u^3v^3(16t^3-8t^2-58t+54)=$$ $$=u^3v^3(t(4t-5)^2+32t^2-83t+54)\geq0$$ because $$83^2-4\cdot32\cdot54=-23<0.$$

2. From the above reasoning we obtain that the maximal value of $k$, for which the inequality $$3(a^2+b^2+c^2) - (a+b+c)^2 \ge \dfrac{k(a-b)^2(b-c)^2(c-a)^2}{a^4+b^4+c^4}$$ is true for any positives $a$, $b$ and $c$ it's $$k_{max}=\min_{t>1}\frac{2(2t^2-1)(2t-1)}{t-1}=25.367...$$

3. I got the following inequality in this type.

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$3(a^2+b^2+c^2) - (a+b+c)^2 \ge \dfrac{48(a-b)^2(b-c)^2(c-a)^2}{a^4+b^4+c^4+5(a^2b^2+a^2c^2+b^2c^2)}.$$

The number $48$ is a best constant.