Find the minimum of the value $k$ such

Question

Let $x\geq0$, $y\geq0$, $z\ge 0$ such that $$\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}\le\sqrt{k(x^2+y^2+z^2)+(6-k)(xy+yz+xz)}$$ Find the minimum of the real value $k$

I square the side $$2(x^2+y^2+z^2)+2\sum_{cyc}\sqrt{(x^2+y^2)(y^2+z^2)}\le k(x^2+y^2+z^2)+(6-k)(xy+yz+xz)$$ $$(k-2)(x^2+y^2+z^2)+(6-k)(xy+yz+xz)\ge 2\sum_{cyc}\sqrt{(x^2+y^2)(y^2+z^2)}$$ Following it seem hard to work

and other hand I think $k_{min}=4\sqrt{2}$,because I let $x=y=1,z=0$,then we have $$2+\sqrt{2}\le\sqrt{2k+(6-k)}\Longrightarrow k\ge 4\sqrt{2}$$

Answer 1

There is a proof of my inequality $$(1+\sqrt2)(x+y+z)\sum_{cyc}\frac{x^2+y^2}{x+y+(\sqrt2-1)z}\leq\sum_{cyc}\left(4\sqrt2x^2+(6-4\sqrt2)xy\right),$$ by SOS.

After full expanding we need to prove that $$\sum_{cyc}((6-4\sqrt2)x^5+(19\sqrt2-26)(x^4y+x^4z)-(15\sqrt2-20)(x^3y^2+x^3z^2)+$$ $$+(94-66\sqrt2)x^3yz-(88-62\sqrt2)x^2y^2z)$$ or $$\sum_{cyc}(6x^5-26x^4y-26x^4z+20x^3y^2+20x^3z^2+94x^3yz-88x^2y^2z)\geq$$ $$\geq\sqrt2\sum_{cyc}(4x^5-19x^4y-19x^4z+15x^3y^2+15x^3z^2+66x^3yz-62x^2y^2z)$$ or $$\sum_{cyc}(3(2x^5-x^4y-x^4z)-23(x^4y+x^4z-x^3y^2-x^3z^2)-3(x^3y^2+x^3z^2-2x^3yz)+$$ $$+88(x^3yz-x^2y^2z))\geq$$ $$\geq\sqrt2\sum_{cyc}(2(2x^5-x^4y-x^4z)-17(x^4y+x^4z-x^3y^2-x^3z^2)-2(x^3y^2+x^3z^2-2x^3yz)+$$ $$+62(x^3yz-x^2y^2z))$$ or $$\sum_{cyc}(x-y)^2((3-2\sqrt2)(x^3+x^2y+xy^2+y^3)+(17\sqrt2-23)xy(x+y)-(3-2\sqrt2)z^3+(44-31\sqrt2)xyz)\geq0$$ and since $3-2\sqrt2>0$, $17\sqrt2-23>0$ and $44-31\sqrt2>0$, it's enough to prove that $$\sum_{cyc}(x-y)^2(x^3+y^3-z^3)\geq0.$$ Now, let $x\geq y\geq z$. Hence, $$\sum_{cyc}(x-y)^2(x^3+y^3-z^3)\geq(x-z)^2(x^3+z^3-y^3)+(y-z)^2(y^3+z^3-x^3)\geq$$ $$\geq(y-z)^2(x^3-y^3)+(y-z)^2(y^3-x^3)=0$$ and we are done!

Answer 2

Yes! For $k=4\sqrt2$ your inequality is true.

Indeed, by C-S $$\left(\sum_{cyc}\sqrt{x^2+y^2}\right)^2\leq\sum_{cyc}(x+y+(\sqrt2-1)z)\sum_{cyc}\frac{x^2+y^2}{x+y+(\sqrt2-1)z}.$$ Thus, it remains to prove that: $$\sum_{cyc}(x+y+(\sqrt2-1)z)\sum_{cyc}\frac{x^2+y^2}{x+y+(\sqrt2-1)z}\leq\sum_{cyc}\left(4\sqrt2x^2+(6-4\sqrt2)xy\right)$$ or $$(1+\sqrt2)(x+y+z)\sum_{cyc}\frac{x^2+y^2}{x+y+(\sqrt2-1)z}\leq\sum_{cyc}\left(4\sqrt2x^2+(6-4\sqrt2)xy\right),$$ which is fifth degree homogeneous inequality.

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, our inequality it's $f(w^3)\geq0$, where $f$ is a linear function,

which says that it's enough to prove the last inequality for an extremal value of $w^3$,

which happens in the following cases.

1. $y=z$.

Since our inequality is homogeneous, we can assume $y=z=1$, which gives $$(x-1)^2x(x+4\sqrt2)\geq0;$$ 2. $w^3=0$.

Let $z=0$ and $y=1$.

We need to prove that $$(x-1)^2(2(\sqrt2-1)^2x^2+5(3\sqrt2-4)x+2(\sqrt2-1)^2)\geq0,$$ which is obvious.

Done!