Given two real numbers $x,y$ so that $x^{2}+y^{2}+xy+4=4y+3x$. Prove that $3\left(x^{3}-y^{3}\right)+20x^{2}+2xy+5y^{2}+39x\leq 100$. 
Given two real numbers $x, y$ so that $x^{2}+ y^{2}+ xy+ 4= 4y+ 3x$. Prove that
$$3\left ( x^{3}- y^{3} \right )+ 20x^{2}+ 2xy+ 5y^{2}+ 39x\leq 100$$

I used derivative and Wolfram|Alpha but only the minimum value found
$$\min\{3\left ( x^{3}- y^{3} \right )+ 20x^{2}+ 2xy+ 5y^{2}+ 39x\}\Leftrightarrow \left\{\begin{matrix} x\cong 0.0320241\\ y\cong 2.16078\\ \left ( z\cong 2.19235 \right ) \end{matrix}\right.$$
where

*

*$z$ is a root of $472z^{3}- 449z^{2}- 689z- 1305= 0$,

*$y$ is a root of $27468y- 11800z^{2}+ 17833z- 41733= 0$,

*$x$ is a root of $27468x+ 3304z^{2}- 11167z+ 7722= 0$.

Why is unsuccessful ? Here are two examples of my claim.

From the condition we obtain:
$$y^2+(x-4)y+x^2-3x+4=0,$$ which gives $$(x-4)^2-4(x^2-3x+4)\geq0$$ or $$0\leq x\leq\frac{4}{3}.$$
By the similar way we obtain: $$1\leq y\leq\frac{7}{3},$$ which gives:
$$3\left ( x^{3}- y^{3} \right )+ 20x^{2}+ 2xy+ 5y^{2}+ 39x\leq$$
$$\leq\frac{64}{9}-3y^3+\frac{320}{9}+\frac{8}{3}y+5y^2+52.$$
Thus, it's enough to prove that:
$$\frac{64}{9}-3y^3+\frac{320}{9}+\frac{8}{3}y+5y^2+52\leq100$$ or
$$9y^3-15y^2-8y+16\geq0$$ or
$$9y^3-24y^2+16y+9y^2-24y+16\geq0$$ or
$$(3y-4)^2(y+1)\geq0$$ and we are done!
The equality occurs for $x=y=\frac{4}{3},$ which says that $100$ is a maximal such value.
