So, there is this problem from Indonesian math olympiad (OSP 2022 no. 3) that bugs me.
We were given that $x$ and $y$ are real numbers that satisfy $$5x^2 + 4xy + 11y^2 = 3$$
Without calculus, find the maximum of
$$xy - 2x + 5y$$
One of the solution is short, but is some kind of black magic. Basically, it goes like this.
$$ \begin{align*} 3 &= 5x^2 + 4xy + 11y^2 \\ &= 5x^2 + 2xy + 2xy + 11y^2 \\ &= 5x^2 + 2xy + 2(k + 2x - 5y) + 11y^2 \\ &= 5x^2 + 2xy + 2k + 4x - 10y + 11y^2 \end{align*} $$
And then somehow it skips to
$$xy - 2x + 5y = \frac{13}{4} - \frac{1}{2}((2x + 1)^2 + (x + y)^2 + 10(y - \frac{1}{2})^2)$$
and it is obvious that $\frac{13}{4}$ is the answer.
What I am asking is not how to find the answer to this problem, rather how or what algorithm to find the form of the sum of square. Is it just by pure luck?
Like... what kind of intuition will help me find this form?
One can brute force or even do something like:
$$k = m - \frac{1}{a}((bx + c)^2 + (cx + e)^2 + z(...)^2)$$
and then match the coefficient. But, I ain't no way solving that ridiculous thing.
Or maybe, first use calculus to get the desired $x$ and $y$, and then set up the form. But, that is some sort of inorganic.
Is there any way to do it? Can AM–GM, Muirhead, Cauchy–Schwarz help?
Without calculus.
From
$$ \cases{ f = x y - 2 x + 5y\\ 5x^2+4 x y+11 y^2= 3 } $$
eliminating $y$ we obtain
$$ 11 f^2+f (4 x+64) x+5 x^4+58 x^3+206 x^2-30 x-75 = 0 $$
now as $f$ is such that is tangent to the restriction we have
$$ 11 f^2+f (4 x+64) x+5 x^4+58 x^3+206 x^2-30 x-75 = (x-x_1)^2(a x^2+ b x + c) $$
there must be a double root $x_1$ due to tangency. Now equating coefficients
$$ \cases{ 5-a = 0\\ 2 a x_1-b+58 = 0\\ -a x_1^2+2 b x_1-c+4 f+206=0\\ -b x_1^2+2 c x_1+64 f-30=0\\ -c x_1^2+11 f^2-75=0 } $$
and with some guess work we obtain two solutions
$$ \left[ \begin{array}{ccccc} f & a & b & c & x_1\\ 10 & 5 & 8 & 41 & -5 \\ \frac{13}{4} & 5 & 53 & \frac{659}{4} & -\frac{1}{2} \\ \end{array} \right] $$