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let $x,y\ge 0$,show that $$4x^3+8y^3+15xy^2-27x-54y+54\ge 0$$

when $x=y=1$ is equality.

see: http://www.wolframalpha.com/input/?i=4x%5E3%2B8y%5E3%2B15xy%5E2-27x-54y%2B54

this inequality is creat by me.and maybe have some methods to prove it? Thank you

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We can prove that in part of the region $x \ge 0, y \ge 0$, the inequality holds$.

Let $f(x,y) = 4x^3+8y^3+15xy^2-27x-54y+54$

If $y=x$ then $$f(x,x)=27(-1 + x)^2(2 + x)\ge 0$$.

Set $x=y+a$ with $a=\frac{3^{3/2}}{2}$, then for $y\ge 0$ we have: $$f(y+a,y)=54 + \frac{81\sqrt{3}}{2}y^2 + 27y^3\ge 0$$

Set $y=x+b$ with $b=\frac{3^{3/2}}{13^{1/2}}$, then for $x\ge 0$ we have: $$f(x,x+b)=\frac{27}{169}\left(338 - 54\sqrt{39} + 78\sqrt{39}x^2 + 169x^3\right)\ge 0$$

This is because $$ 338^2=114244>113724= (54\sqrt{39})^2$$.

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Find the point/s where the gradient vanishes, check that these evaluate to $\geq 0$, and show these are minimas. In case of saddle points, show these evaluate to $>0$.

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  • $\begingroup$ what? I want to see such AM-GM,sos,Cauchy-Schwarz ,Holder,and so on to prove it.. $\endgroup$ – math110 Jun 21 '14 at 15:41
  • $\begingroup$ @math110 Perhaps you want to state this in the question. Feel free to down vote :) $\endgroup$ – user76568 Jun 21 '14 at 15:45
  • $\begingroup$ Also check along $x=0$ and $y=0$, where it is $(2y-3)^2(2y+6)$ and $(2x-3)^2(x+3)+27$. $\endgroup$ – Empy2 Jun 21 '14 at 16:34
  • $\begingroup$ I get a saddle point at $(x,y)=(\sqrt{3/28},\sqrt{12/7})$. $\endgroup$ – Empy2 Jun 21 '14 at 16:56
  • $\begingroup$ @Michael Correct, but it evaluates to a positive number, and hence $(1,1)$ is the only minimum for $x,y \geq 0$ $\endgroup$ – user76568 Jun 21 '14 at 17:39
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Change variables to $x=X+1,y=Y+1$, and I think you get $$4X^3+12X^2+8Y^3+39Y^2+15XY^2+30XY$$

which is positive if $X$ and $Y$ are positive - so if $x\geq 1,y\geq 1$.

If $X\geq 0$, then it is at least $12X^2+30XY+31Y^2$, which is always positive.

When $x<1,y<1$, then $F_x$ and $F_y$ are both negative, so $F(x,y)>F(1,1)$.

When $y>3/2$, $F_y(x,y)=24y^2+30xy-54>0$, so $F(x,y)>F(x,3/2)$

I still have $0<x<1,1<y<3/2$ to go.

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