# Find minimum value of $\frac{x^3}{\sqrt{2(y^4+1)}}+\frac{y^3}{\sqrt{2(z^4+1)}}+\frac{z^3}{\sqrt{2(x^4+1)}}$

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

• You can use write $0 \leqslant \frac{2}{9}{(x + y + z)^3} - {x^4} + {y^4} + {z^4} + 3$ as a sum of square then you can proof your inequality. To do this please look at this post math.stackexchange.com/questions/2410994/… Aug 4, 2021 at 7:03
• @TuongNguyenMinh Thank you, i'll try it Aug 4, 2021 at 10:12
• It should be noted that these inequalities from IMO exam can be tackle quite easily using s.o.s decomposition due to their symmetrical form. Aug 4, 2021 at 15:48

We have \begin{align*} 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)}}\\ &\ge \frac{x^3}{2(y^2 - y + 1)} + \frac{y^3}{2(z^2 - z + 1)} + \frac{z^3}{2(x^2 - x + 1)} \tag{1}\\ &\ge \frac{(x^2 + y^2 + z^2)^2}{2x(y^2 - y + 1) + 2y(z^2 - z + 1) + 2z(x^2 - x + 1)} \tag{2}\\ &= \frac{(x^2 + y^2 + z^2)^2}{2(xy^2 + yz^2 + zx^2) + 2(x + y + z - xy - yz - zx)}\\ &= \frac{(x^2 + y^2 + z^2)^2}{2(xy^2 + yz^2 + zx^2)}\\ &\ge \frac{(x^2 + y^2 + z^2)^2}{2\sqrt{(x^2y^2 + y^2z^2 + z^2x^2)(y^2 + z^2 + x^2)}} \tag{3}\\ &\ge \frac{(x^2 + y^2 + z^2)^2}{2\sqrt{\frac{(x^2 + y^2 + z^2)^2}{3}(y^2 + z^2 + x^2)}} \tag{4}\\ &= \frac{1}{2}\sqrt{3(x^2 + y^2 + z^2)}\\ &\ge \frac{1}{2}\sqrt{3(xy + yz + zx)}\\ &\ge \frac{3}{2}. \tag{5} \end{align*} Explanations:
(1) the identity $$4(u^2 - u + 1)^2 = 2(u^4 + 1) + 2(u - 1)^4$$;
(2) the Cauchy-Bunyakovsky-Schwarz inequality;
(3) the Cauchy-Bunyakovsky-Schwarz inequality;
(4) $$(a + b + c)^2 \ge 3(ab + bc + ca)$$ for all reals $$a, b, c$$;
(5) $$(xy + yz + zx)^2 = (x + y + z)^2 \ge 3(xy + yz + zx)$$, so $$xy + yz + zx \ge 3$$.

Also, if $$x = y = z = 1$$, then $$P = 3/2$$. Thus, the minimum of $$P$$ is $$3/2$$.

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?

• Thank you sir, but i'm new at inequality so i can't use Holder, btw thank you, I will try sum of square. Aug 4, 2021 at 10:11