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let $x^3+y^3+z^3=3,x,y,z>0$ show that $$\dfrac{x}{y^2+5}+\dfrac{y}{z^2+5}+\dfrac{z}{x^2+5}\le\dfrac{1}{2}$$

I have show that let $x,y,z$ be positive numbers,such that $x+y+z=3$,prove that $$\dfrac{x}{1+y^3}+\dfrac{y}{1+z^3}+\dfrac{z}{1+x^3}\ge\dfrac{3}{2}$$ pf: use $AM-GM$ we have $$\dfrac{x}{1+y^3}=x-\dfrac{xy^3}{1+y^3}\ge x-\dfrac{xy^3}{2y^{3/2}}=x-\dfrac{xy^{3/2}}{2}$$ and,similarly $$\dfrac{y}{1+z^3}\ge y-\dfrac{yz^{3/2}}{2},\dfrac{z}{1+x^3}\ge z-\dfrac{zx^{3/2}}{2}$$ Thus,it suffices to show that $$xy^{3/2}+yz^{3/2}+zx^{3/2}\le 3$$ and it is known that $$(a^3b^2+b^3c^2+c^3a^2)^2\le\dfrac{1}{3}(a^2+b^2+c^2)^3$$ seting $x=a^2,y=b^2,z=c^2$,by done!

But for this problem : $$y^2+5=y^2+1+1+1+1+1\ge 6y^{1/3}$$ and similarly $$z^2+5\ge 6z^{1/3}, x^2+5\ge 6x^{1/3}$$ it suffices prove that $$xy^{-1/3}+yz^{-1/3}+zx^{-1/3}\le 3$$ with $x^3+y^3+z^3=3$,I use maple find this is ($xy^{-1/3}+yz^{-1/3}+zx^{-1/3}\le 3$,with $x^3+y^3+z^3=3$) not true!,But after I use maple find this $\dfrac{x}{y^2+5}+\dfrac{y}{z^2+5}+\dfrac{z}{x^2+5}\le\dfrac{1}{2},x^3+y^3+z^3=3$ is true!

and my other idea $$\dfrac{x}{y^2+5}+\dfrac{y}{z^2+5}+\dfrac{z}{x^2+5}\le\dfrac{1}{2}$$ $$\Longleftrightarrow 2\sum x^2y^3+10\sum x^2y+50\sum x-5\sum x^2y^2-25\sum y^2-x^2y^2z^2\le 95$$

so I think my methods can't prove this problem, can someone use other methods show it? Thank you everyone.

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marked as duplicate by Ewan Delanoy, Start wearing purple, Micah, Jim, rschwieb Jun 13 '13 at 17:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I don't really undestand, what do you want. Just to solve this inequality? – Harold Jun 13 '13 at 13:41
up vote 2 down vote accepted

By Cauchy-Schwarz inequality $$\frac{x}{y^2 + 5} + \frac{y}{z^2 + 5} + \frac{z}{x^2+5} \le \sqrt{x^2 + y^2 + z^2}\sqrt{\frac{1}{(x^2+5)^2} + \frac{1}{(y^2 + 5)^2} + \frac{1}{(z^2 + 5)^2}}.$$ By AM-GM the RHS is less than $$\frac{\frac{x^2 + y^2 + z^2}{6} + \frac{6}{(x^2 +5)^2} + \frac{6}{(y^2+5)^2} + \frac{6}{(z^2+5)^2}}{2} = \sum \left(\frac{x^2}{12} + \frac{3}{(x^2+5)^2}\right).$$ By multiplying everything out and factoring, one can prove that for $x \in [0,\sqrt[3]{3}]$ $$\frac{x^2}{12} + \frac{3}{(x^2+5)^2} \le \frac{2x^3 + 7}{54}.$$ Indeed, the inequality is equivalent with $$(x-1)^2(4x^5 - x^4 + 34x^3 - 7x^2 + 52x + 26) \ge 0,$$ which is clearly true for $x \ge 0$. After we have this result, we get that $$\frac{x}{y^2 + 5} + \frac{y}{z^2 + 5} + \frac{z}{x^2+5} \le \frac{2(x^3+y^3+z^3) + 3\cdot 7}{54} = \frac{1}{2}.$$

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Thank you, Why $$4x^5-x^4+34x^3-7x^2+52x+26\ge 0,0<x\le \sqrt[3]{3}?$$ – math110 Jun 13 '13 at 16:55
@math110: When $0 \le x \le 1$, $26 \ge x^4 + 7x^2$ and when $x \ge 1$, $4x^5 + 34x^3 \ge x^4 + 7x^2$. – J. J. Jun 13 '13 at 17:00
Nice,Thank you @J.J – math110 Jun 14 '13 at 3:36
Thank you,can you solution my this problem? Thank… – math110 Jun 14 '13 at 5:24

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