For $x=y=z$ the inequality $\sum\limits_{cyc}\frac{x^2}{\sqrt{x^2+y^2}}\geq k(x+y+z)$ gives $k\leq\frac{1}{\sqrt2}$.
The Peter Scholze's solution for $k=\frac{1}{\sqrt2}.$:
By Rearrangement $$\sum_{cyc}\frac{x^2}{\sqrt{x^2+y^2}}=\sqrt{\sum_{cyc}\left(\frac{x^4}{x^2+y^2}+\frac{2x^2y^2}{\sqrt{x^2+y^2}}\cdot\frac{1}{\sqrt{y^2+z^2}}\right)}\geq$$
$$\geq\sqrt{\sum_{cyc}\left(\frac{x^4}{x^2+y^2}+\frac{2x^2y^2}{\sqrt{x^2+y^2}}\cdot\frac{1}{\sqrt{x^2+y^2}}\right)}=$$
$$=\sqrt{\sum_{cyc}\left(\frac{x^4}{x^2+y^2}+\frac{2x^2y^2}{x^2+y^2}\right)}\geq\frac{x+y+z}{\sqrt2},$$ where the last inequality it's just $$\sum_{cyc}\frac{(x-y)^4}{x^2+y^2}\geq0.$$
In the making of Rearrangement we used the following reasoning.
The triples $\left(\frac{x^2y^2}{\sqrt{x^2+y^2}},\frac{x^2z^2}{\sqrt{x^2+z^2}},\frac{y^2z^2}{\sqrt{y^2+z^2}}\right)$ and $\left(\frac{1}{\sqrt{x^2+y^2}},\frac{1}{\sqrt{x^2+z^2}},\frac{1}{\sqrt{y^2+z^2}}\right)$ have the opposite ordering, which gives a possibility to use Rearrangement.
I think, TL method does not help here.
Another way.
Let $a=x^3$, $b=y^3$ and $c=z^3$.
Thus, $x^3+y^3+z^3=3$ and we need to prove that:
$$x+y+z\geq x^3y^3+x^3z^3+y^3z^3$$ or
$$(x+y+z)^3(x^3+y^3+z^3)^5\geq243(x^3y^3+x^3z^3+y^3z^3)^3.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, we need to prove that $f(w^3)\geq0,$ where
$$f(w^3)=u^3(9u^3-9uv^2+w^3)^5-(9v^6-9uv^2w^3+w^6)^3.$$
But, $$f'(w^3)=5u^3(9u^3-9uv^2+w^3)^4+2(9uv^2-2w^3)(9v^6-9uv^2w^3+w^6)^2\geq0,$$ which says that $f$ increases and it's enough to prove our inequality for the minimal value of $w^3$.
Now, $x$, $y$ and $z$ are positive roots of the equation
$$(t-x)(t-y)(t-z)=0$$ or
$$t^3-3ut^2+3v^2t-w^3=0,$$ which says that a graph of $f(t)=t^3-3ut^2+3v^2t$ and the line $g(t)=w^3$ have three common points.
But $$f'(t)=3t^2-6ut+3v^2=3(t-(u+\sqrt{u^2-v^2}))(t-(u-\sqrt{u^2-v^2})),$$
which gives $t_{min}=u+\sqrt{u^2-v^2}$ and since $f(0)=0$, we obtain two cases:
- $f\left(t_{min}\right)>0$.
In this case $w^3$ gets a minimal value, when the line $g(t)=w^3$ is a tangent line to the graph of $f$, which says that in this case it's enough to prove our inequality for equality case of two variables.
We'll prove this for the inequality $$(x+y+z)^3(x^3+y^3+z^3)^5\geq243(x^3y^3+x^3z^3+y^3z^3)^3.$$
It's enough to assume $y=z=1$ and we need to prove that:
$$(x+2)^3(x^3+2)^5\geq243(2x^3+1)^3$$ or $h(x)\geq0,$ where
$$h(x)=3\ln(x+2)+5\ln(x^3+2)-3\ln(2x^3+1)-5\ln3.$$
But $$h'(x)=\frac{(x-1)(3x^5+7x^4+7x^3+6x^2-x-1)}{(x+2)(x^3+2)(2x^3+1)},$$
which gives $x_{min}=1$, $x_{max}=0.375...$ and since $h(1)=0$ it's enough to prove that $h(0)\geq0$ or $256\geq243,$ which is true.
- $w^3\rightarrow0^+$.
Let $z\rightarrow0^+$ and $y=1$.
Thus, we need to prove that $$(x+1)^3(x^3+1)^5\geq243x^9,$$ which is true by AM-GM because $256>243.$
Best Answer
Originally a comment and similar to Gae. S.:
You could say $$\begin{matrix} & 121 &>& 108 &>& 100 \\ \implies& \sqrt{121} &>& \sqrt{108} &>& \sqrt{100}\\ \implies& 11 &>& 6 \sqrt{3} &>& 10\\ \implies& -11 &<& - 6 \sqrt{3} &<& -10\\ \implies &0 &<& 11- 6 \sqrt{3} &<& 1\end{matrix}$$