Prove that $\sqrt{a^2+3b^2}+\sqrt{b^2+3c^2}+\sqrt{c^2+3a^2}\geq6$ if $(a+b+c)^2(a^2+b^2+c^2)=27$

Question

Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b+c)^2(a^2+b^2+c^2)=27$. Prove that: $$\sqrt{a^2+3b^2}+\sqrt{b^2+3c^2}+\sqrt{c^2+3a^2}\geq6$$

A big problem here around $(a,b,c)=(1.6185...,0.71686...,0.4926...)$.

In this case we get $\sum\limits_{cyc}\sqrt{a^2+3b^2}-6=0.000563...$.

My trying.

Let $a^2+3b^2=4x^2$, $b^2+3c^2=4y^2$ and $c^2+3a^2=4z^2$, where $x$, $y$ and $z$ are non-negatives.

Hence, we need to prove that $$\sum\limits_{cyc}\sqrt{x^2-3y^2+9z^2}\leq\frac{\sqrt7(x+y+z)^2}{\sqrt{3(x^2+y^2+z^2)}}$$

Let $k$ and $m$ be non-negatives, for which

$x-ky+mz>0$, $y-kz+mx>0$, $z-kx+my>0$ and $1-k+m>0$.

By C-S $\left(\sum\limits_{cyc}\sqrt{x^2-3y^2+9z^2}\right)^2\leq(1-k+m)(x+y+z)\sum\limits_{cyc}\frac{x^2-3y^2+9z^2}{x-ky+mz}$.

Thus, it remains to prove that $$(1-k+m)\sum\limits_{cyc}\frac{x^2-3y^2+9z^2}{x-ky+mz}\leq\frac{7(x+y+z)^3}{3(x^2+y^2+z^2)}$$

It's a sixth degree, but I didn't find a values of $k$ and $m$, such that the last inequality will be true.

By this way we can prove that $\sum\limits_{cyc}\sqrt{a^2+2b^2}\geq3\sqrt3$ is true, but it's not comforting.

Also I tried to use Holder, but without success.

Thank you!

Answer 1

There seems to be bugs in the segment after where I marked "***". Needed for check.
When I saw the form $\sqrt{a^2+3b^2}$ I thought of the absolute value of a complex number.
So let $$u=a+\sqrt{3}bi$$ $$v=b+\sqrt{3}ci$$ $$w=c+\sqrt{3}ai$$ And now what you want to prove becomes $$|u|+|v|+|w|\geq6$$

$$u+v+w=(1+\sqrt{3}i)(a+b+c)$$ $$|u|^2+|v|^2+|w|^2=4(a^2+b^2+c^2)$$
$$(u+v+w)^2(|u|^2+|v|^2+|w|^2)$$ $$=(1+\sqrt{3}i)^2(a+b+c)^2\cdot4(a^2+b^2+c^2)$$ $$=4(1+\sqrt{3}i)^2(a+b+c)^2(a^2+b^2+c^2)$$ $$=4(1+\sqrt{3}i)^2\cdot27$$
$$|u+v+w|^2(|u|^2+|v|^2+|w|^2)=|4(1+\sqrt{3}i)^2\cdot27|=4\cdot27\cdot4$$ Now I thought I should separate $|u+v+w|$ to $|u|+|v|+|w|$ so that all the elements in the expression were independent $|u|$, $|v|$, $|w|$, and its form would be closer to the inequality we want to prove.
So I used the triangle inequality, $$|u|+|v|+|w|\geq|u+v|+|w|\geq|u+v+w|$$ $$(|u|+|v|+|w|)^2(|u|^2+|v|^2+|w|^2)\geq|u+v+w|^2(|u|^2+|v|^2+|w|^2)=4\cdot27\cdot4$$ Let $x=|u|,\ \ y=|v|,\ \ z=|w|$
Then the problem becomes, $$\mathrm{if\ \ }(x+y+z)^2(x^2+y^2+z^2)\geq4\cdot27\cdot4$$ $$\mathrm{prove\ that\ \ }x+y+z\geq6$$ Proof:
let $k$ be a number so that $x+y+z\geq k\geq0$ is always true.
A known formula: $$(x-y)^2+(y-z)^2+(z-x)^2=3(x^2+y^2+z^2)-(x+y+z)^2\geq0$$ ***Then $$3(x^2+y^2+z^2)\geq(x+y+z)^2\geq k^2$$ $$\implies (x+y+z)^2(x^2+y^2+z^2)\geq k^2\cdot\frac{k^2}{3}=\frac{k^4}{3}$$ But it's already known that $(x+y+z)^2(x^2+y^2+z^2)\geq4\cdot27\cdot4$ has to be true. So to let $$(x+y+z)^2(x^2+y^2+z^2)\geq\frac{k^4}{3}$$ be always true, $$\frac{k^4}{3}\leq4\cdot27\cdot4$$ $$k\leq6$$ This proof doesn't need $a,b,c\geq0$. It just needs them to be real numbers.