$\sqrt{2(x^2+y^2)}+\sqrt{2(y^2+z^2)}+\sqrt{2(z^2+x^2)}\ge \sqrt[3]{9(x+y)^3+9(y+z)^3+9(z+x)^3}$?

Question

Let $x,y,z>0$. Show that $$\sqrt{2(x^2+y^2)}+\sqrt{2(y^2+z^2)}+\sqrt{2(z^2+x^2)}\ge \sqrt[3]{9(x+y)^3+9(y+z)^3+9(z+x)^3}$$

I have tried squaring and Cauchy-Schwarz and other inequalities, all not solves it.

Answer 1

By C-S $$\sum_{cyc}\sqrt{x^2+y^2}=\sqrt{2(x^2+y^2+z^2)+2\sum_{cyc}\sqrt{(x^2+y^2)(x^2+z^2)}}=$$ $$=\sqrt{2(x^2+y^2+z^2)+2\sqrt{\sum_{cyc}\left(x^4+3x^2y^2+2(x^2+y^2)\sqrt{(x^2+z^2)(y^2+z^2)}\right)}}\geq$$ $$\geq\sqrt{2(x^2+y^2+z^2)+2\sqrt{\sum_{cyc}\left(x^4+3x^2y^2+2(x^2+y^2)(xy+z^2)\right)}}=$$ $$=\sqrt{2(x^2+y^2+z^2)+2\sqrt{\sum_{cyc}(x^4+2x^3y+2x^3z+7x^2y^2)}}.$$ Thus, it remains to prove that $$2\sqrt{x^2+y^2+z^2+\sqrt{\sum_{cyc}(x^4+2x^3y+2x^3z+7x^2y^2)}}\geq\sqrt[3]{9\sum_{cyc}(2x^3+3x^2y+3x^2z)}.$$ Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, we need to prove that $$2\sqrt{9u^2-6v^2+\sqrt{81u^4-54u^2v^2+45v^4-36uw^3}}\geq3\sqrt[3]{18u^3-9uv^2-w^3}$$ or $f(w^3)\geq0$, where $$f(w^3)=2\sqrt{9u^2-6v^2+\sqrt{81u^4-54u^2v^2+45v^4-36uw^3}}-3\sqrt[3]{18u^3-9uv^2-w^3}.$$ Now,$$f'(w^3)=\frac{\frac{-18u}{\sqrt{81u^4-54u^2v^2+45v^4-36uw^3}}}{\sqrt{9u^2-6v^2+\sqrt{81u^4-54u^2v^2+45v^4-36uw^3}}}+\frac{1}{\sqrt[3]{(18u^3-9uv^2-w^3)^2}}=$$ $$=\frac{1}{\sqrt[3]{(18u^3-9uv^2-w^3)^2}}-$$ $$-\frac{6u}{\sqrt{9u^4-6u^2v^2+5v^4-4uw^3}\sqrt{9u^2-6v^2+3\sqrt{9u^4-6u^2v^2+5v^4-4uw^3}}}\leq$$ $$\leq\frac{1}{\sqrt[3]{(18u^3-10uv^2)^2}}-\frac{6u}{3u^2\cdot\sqrt{9u^2+9u^2}}\leq0.$$ Thus, $f$ is decreasing function,

which says that it's enough to prove $f(w^3)\geq0$ for a maximal value of $w^3$.

We know that $x$, $y$ and $z$ are roots of the equation $$(X-x)(X-y)(X-z)=0$$ or $$X^3-3uX^2+3v^2X-w^3=0$$ or $$X^3-3uX^2+3v^2X=w^3,$$ which says that a graph of function $F(X)=X^3-3uX^2+3v^2X$ and the line $Y=w^3$ have three common points.

Let now $w^3$ increases and $u$ and $v^2$ do not change.

We see that $w^3$ will get a maximal value, when a line $Y=w^3$ is a tangent line to graph of $F$,

which happens for equality case of two variables.

Since $f(w^3)\geq0$ is homogeneous inequality, we can assume $y=z=1$ and we need to prove that $$2\sqrt{x^2+2+\sqrt{x^4+4x^3+14x^2+4x+13}}\geq\sqrt[3]{18(x^3+3x^2+3x+5)}$$ and the rest is smooth.