# How Prove $4x^3+8y^3+15xy^2-27x-54y+54\ge 0$

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

when $x=y=1$ is equality.

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

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 $$338^2=114244>113724= (54\sqrt{39})^2$$. 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$>0$. • what? I want to see such AM-GM,sos,Cauchy-Schwarz ,Holder,and so on to prove it.. – math110 Jun 21 '14 at 15:41 • @math110 Perhaps you want to state this in the question. Feel free to down vote :) – user76568 Jun 21 '14 at 15:45 • Also check along$x=0$and$y=0$, where it is$(2y-3)^2(2y+6)$and$(2x-3)^2(x+3)+27$. – Empy2 Jun 21 '14 at 16:34 • I get a saddle point at$(x,y)=(\sqrt{3/28},\sqrt{12/7})$. – Empy2 Jun 21 '14 at 16:56 • @Michael Correct, but it evaluates to a positive number, and hence$(1,1)$is the only minimum for$x,y \geq 0$– user76568 Jun 21 '14 at 17:39 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.