# Proof of a tighter inequality than Cauchy-Schwarz inequality

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?

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 $$83^2-4\cdot32\cdot54=-23<0.$$
1. 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 $$k_{max}=\min_{t>1}\frac{2(2t^2-1)(2t-1)}{t-1}=25.367...$$
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.