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let $x,y,z>0$, and $x+3y+z=9$, find the minimum of

$$x+y^2+z^3$$

I think this problem is very interesting. I have found this when $$x=\dfrac{9}{2}-\dfrac{1}{\sqrt{3}},y=\dfrac{3}{2},z=\dfrac{1}{\sqrt{3}}$$

I belive this inequality have $AM-GM$ methods,becasue I have see this same problem can use $AM-GM$ methods,and I think this methods is very very nice.

if $a,b,c>0$,and $ a+b^2+c^3=\dfrac{325}{9}$, prove that $$a^2+b^3+c^4\ge\dfrac{2807}{27}$$

my methods: let $a=x,b=y,c=z$,and then $$x+y^2+z^3=\dfrac{325}{9}$$ use $AM-GM$,we have $$a^2+x^2\ge 2ax$$ $$b^3+b^3+y^3\ge 3yb^2$$ $$c^4+c^4+c^4+z^4\ge 4zc^3$$

then we have $$a^2+b^3+c^4+x^2+\dfrac{y^3}{2}+\dfrac{z^4}{3}\ge 2ax+\dfrac{3}{2}b^2+\dfrac{4}{3}zc^3$$

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    $\begingroup$ Did you use Lagrange multiplier method? $\endgroup$ – Mhenni Benghorbal May 14 '13 at 11:35
  • $\begingroup$ I don't use Lagrange multiplier $\endgroup$ – math110 May 14 '13 at 11:35
  • $\begingroup$ Your solution $x=\dfrac{9}{2}-\dfrac{1}{\sqrt{3}},y=\dfrac{3}{2},z=\dfrac{1}{\sqrt{3}}$ is correct. $\endgroup$ – Mhenni Benghorbal May 14 '13 at 11:52
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Since $x+y^2+a^2+z^3+b^3+b^3\ge x+2ay+3b^2z$, now let $2a=3, 3b^2=1$. You can determine $ y=a=\frac{3}{2}, z=b=\frac{1}{\sqrt{3}}$ and $x=9-3y-z$ if equality holds.

You do get an AM-GM approach.

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  • $\begingroup$ Thank you ,the same as me. $\endgroup$ – math110 May 14 '13 at 16:08

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