Let $x,y,z>$ and $x+y+z=xy+yz+zx$
. Find the minimum value of $$P=\frac{x^3}{\sqrt{2(y^4+1)}}+\frac{y^3}{\sqrt{2(z^4+1)}}+\frac{z^3}{\sqrt{2(x^4+1)}}$$
My solution: I know the minimum value is $\frac{3}{2}$ when $x=y=z=1$
So $$P=\frac{x^4}{\sqrt{x^2.2(y^4+1)}}+\frac{y^4}{\sqrt{y^2.2(z^4+1)}}+\frac{z^4}{\sqrt{z^2.2(x^4+1)}}$$
$$\ge\frac{(x^2+y^2+z^2)^2}{\sqrt{2x^2(y^4+1)}+\sqrt{2y^2(z^4+1)}+\sqrt{2z^2(x^4+1)}}$$
$$\ge\frac{(x^2+y^2+z^2)^2}{\sqrt{2(x^2+y^2+z^2)(y^4+1+z^4+1+x^4+1)}}$$
$$\ge\sqrt{\frac{(x+y+z)^3}{2(x^4+y^4+z^4+3)}}$$ because $(x^2+y^2+z^2)\ge(x+y+z)$
So now i need prove $x^4+y^4+z^4+3 \le \frac{2}{9}(x+y+z)^3$ but i stuck here for a hour. So please help me, thnank
Best Answer
The inequality, which you got is wrong. Try $x=y=2$ and $z\rightarrow0^+$.
For $x=y=z=1$ we obtain a value $\frac{3}{2}$.
We'll prove that it's a minimal value.
Indeed, by Holder $$\sum_{cyc}\frac{x^3}{\sqrt{y^4+1}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{x^3}{\sqrt{y^4+1}}\right)^2\sum\limits_{cyc}x^3(y^4+1)}{\sum\limits_{cyc}x^3(y^4+1)}}\geq\sqrt{\frac{(x^3+y^3+z^3)^3}{\sum\limits_{cyc}x^3(y^4+1)}}$$ and it's enough to prove that: $$2(x^3+y^3+z^3)^3\geq9\sum_{cyc}(x^4z^3+x^3),$$ which is true because $$(x^3+y^3+z^3)^3\geq9\sum_{cyc}x^4z^3$$ and $$(x^3+y^3+z^3)^2\geq9.$$ Can you end it now?