I propose a little refinement of this an difficult inequality from Crux mathematicorum .
Claim:
Let $a,b,c>0$ such that $a\geq b\geq c$ and $\frac{a}{c}\geq\frac{b}{a}\geq \frac{c}{b}$ and $b\geq a\frac{(b+c)}{(a+b)}$ then we have :
$$\sqrt{a^2+3b^2}+\sqrt{b^2+3c^2}+\sqrt{c^2+3a^2}\geq \sqrt{3(c+b)^2+(a+b)^2}+\sqrt{c^2+3a^2}\geq \sqrt{7(a^2+b^2+c^2)+5(ab+bc+ca)}$$
Proof :
First we show :
$$\sqrt{a^2+3b^2}+\sqrt{b^2+3c^2}+\sqrt{c^2+3a^2}\geq \sqrt{3(c+b)^2+(a+b)^2}+\sqrt{c^2+3a^2}$$
Rewriting the inequality :
$$a\Bigg(\sqrt{1+3\Big(\frac{b^2}{a^2}\Big)}\Bigg)+b\Bigg(\sqrt{1+3\Big(\frac{c^2}{b^2}\Big)}\Bigg)+c\Bigg(\sqrt{1+3\Big(\frac{a^2}{c^2}\Big)}\Bigg)\geq (a+b)\Bigg(\sqrt{1+3\Bigg(\frac{(b+c)^2}{(a+b)^2}\Bigg)}\Bigg)+c\Bigg(\sqrt{1+3\Big(\frac{a^2}{c^2}\Big)}\Bigg)$$
We introduce the function wich is convex,increasing on $(0,\infty)$:
$$f(x)=\sqrt{1+3x^2}$$
We have in fact :
$$af\Big(\frac{b}{a}\Big)+bf\Big(\frac{c}{b}\Big)+cf\Big(\frac{a}{c}\Big)\geq af\Bigg(\frac{(b+c)}{(a+b)}\Bigg)+bf\Bigg(\frac{(b+c)}{(a+b)}\Bigg)+cf\Big(\frac{a}{c}\Big)\quad (1)$$
As the function is convex we use weighted Karamata's inequality to prove $(1)$ :
The majorization proceeds as following :
$$c\Big(\frac{a}{c}\Big)=c\Big(\frac{a}{c}\Big)$$
$$c\Big(\frac{a}{c}\Big)+a\Big(\frac{b}{a}\Big)\geq c\Big(\frac{a}{c}\Big)+a\frac{(b+c)}{(a+b)}$$
$$c\Big(\frac{a}{c}\Big)+a\Big(\frac{b}{a}\Big)+b\Big(\frac{c}{b}\Big)=c\Big(\frac{a}{c}\Big)+a\frac{(b+c)}{(a+b)}+b\frac{(b+c)}{(a+b)}$$
So the majorization is complete we can apply weighted Karamata's inequality and we are done with $(1)$
Now we show :
$$\sqrt{3(c+b)^2+(a+b)^2}+\sqrt{c^2+3a^2}\geq \sqrt{7(a^2+b^2+c^2)+5(ab+bc+ca)}\quad (2)$$
After some basic manipulations and using Buffalo's way it's true because :
$$12x^2z^2-12xz^3+3z^4=z^2(3(z-2x)^2)\geq 0$$
My question
First it is right ?
Have you an alternative proof of $(2)$ so without Buffalo's way?
Thanks in advance !
Regards Max.