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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

Michael Rozenberg
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tompi2394
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    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

    https://math.stackexchange.com/questions/2410994/write-x2-y2-z22-3-x3-y-y3-z-z3-x-as-a-sum-of-three-squ/2414847#2414847

     – Tuong Nguyen Minh Aug 04 '21 at 07:03
  • @TuongNguyenMinh Thank you, i'll try it – tompi2394 Aug 04 '21 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. – Tuong Nguyen Minh Aug 04 '21 at 15:48

2 Answers2

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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?

Michael Rozenberg
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  • Thank you sir, but i'm new at inequality so i can't use Holder, btw thank you, I will try sum of square. – tompi2394 Aug 04 '21 at 10:11
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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 $3/2$.

River Li
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