<s>[INST] 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.
 [/INST] Alternative solution
$x^2 + y^2 + xy + 4 = 4y + 3x$ is written as
$$(y + \tfrac{x}{2} - 2)^2 = \tfrac{1}{4}x(4-3x).\tag{1}$$
From $\frac{1}{4}x(4-3x) \ge 0$, we have $0 \le x \le \frac{4}{3}$.
If $x = 0$, then $y = 2$ and $3(x^3 - y^3) + 20x^2 + 2xy + 5y^2 + 39x = -4 < 100$.
If $0 < x \le \frac{4}{3}$, let $x = \frac{4}{t^2 + 3}$. Then from (1), we have
$$y = \frac{2t^2 - 2t + 4}{t^2 + 3}, \quad \mathrm{or}, \quad y = \frac{2t^2 + 2t + 4}{t^2 + 3}.$$
Denote $F = 3(x^3 - y^3) + 20x^2 + 2xy + 5y^2 + 39x$. There are two possible cases:

*

*If $y = \frac{2t^2 - 2t + 4}{t^2 + 3}$, we have
$$F = \frac{-4t^4+32t^3+128t^2+900}{(t^2+3)^2}$$
and
$$100 - F = \frac{8t^2(13t^2-4t+59)}{(t^2+3)^2} \ge 0.$$


*If $y = \frac{2t^2 + 2t + 4}{t^2 + 3}$, we have
$$F = \frac{-4t^4-32t^3+128t^2+900}{(t^2+3)^2}$$
and
$$100 - F = \frac{8t^2(13t^2 + 4t + 59)}{(t^2+3)^2} \ge 0.$$
We are done.
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