I met an equation in a paper: \begin{equation} \mu^2(\mu^2-rs)=(\mu^2-rs)(a+b)+2ab\mu+a^2s+b^2r \end{equation} we already have known $\mu^2<rs$ and $r>0$.In the paper, the author gives a magical substitution $\alpha=2a-r+\mu,\beta=2b-s+\mu$ and then: \begin{equation} r(r+s-2\mu-4\mu^2)(-\mu^2+rs)=(-\mu^2+rs)\alpha^2+(r\beta+\mu\alpha)^2 \end{equation} directly implying that $r+s-2\mu-4\mu^2\geq 0$.
Here are my questions:
- what is the motivation for this substitution? I think more specific details will help a lot.
- Is there such a technique to deform the equation which is always positive into a sum of squares?
Thanks for your help!
Remark: Finding $h$ such that $f + h\cdot g \ge 0$ is a trick in proving some inequalities.
Problem: Let $a, b, \mu, s, r$ be real numbers such that $r > 0$ and $\mu^2 < rs$ and $\mu^2(\mu^2-rs)=(\mu^2-rs)(a+b)+2ab\mu+a^2s+b^2r$. Prove that $r+s-2\mu-4\mu^2\geq 0$.
Proof:
Let $$f := r+s-2\mu-4\mu^2$$ and $$g := (\mu^2-rs)(a+b)+2ab\mu+a^2s+b^2r - \mu^2(\mu^2-rs).$$
We hope to find a function $h$ such that $$f + h\cdot g \ge 0.$$ (Note: This implies $f \ge 0$.)
We chose $$h = \frac{4}{rs - \mu^2}.$$ (See Remarks at the end for details.)
Note that $f + \frac{4}{rs - \mu^2}\cdot g$ is quadratic in $b$. It is easy to get \begin{align*} f + \frac{4}{rs - \mu^2}\cdot g &= {\frac { \left( 2\,a\mu +2\,br-rs+{\mu}^{2} \right) ^{2}}{r(rs-\mu^2)}}+{\frac { \left( 2\,a-r+\mu \right) ^{2} }{r}}\\ &= \frac{(r\beta + \mu \alpha)^2}{r(rs- \mu^2)} + \frac{\alpha^2}{r} \end{align*} or $$r\cdot f \cdot (rs - \mu^2) + 4r \cdot g = (rs - \mu^2)\alpha^2 + (r\beta + \mu \alpha)^2.$$ In other words, given that $g = 0$, we have $$r\cdot f \cdot (rs - \mu^2) = (rs - \mu^2)\alpha^2 + (r\beta + \mu \alpha)^2$$ which is exactly the result in the paper.
We are done.
Remarks:
Note that $f + h\cdot g$ is quadratic in $b$, i.e. $$f + h\cdot g = q_1 b^2 + q_2b + q_3$$ where \begin{align*} q_1 &= r h, \\ q_2 &= 2\,ah\mu-hrs+h{\mu}^{2}, \\ q_3 &= hrs{\mu}^{2}-h{\mu}^{4}+{a}^{2}hs-ahrs+ah{\mu}^{2}-4\,{\mu}^{2}+r+s-2\,\mu. \end{align*} We hope to find $h$ such that $q_1 > 0$ and $4q_1q_3 \ge q_2^2$. Note that $4q_1q_3 - q_2^2$ is quadratic in $a$, i.e. $$ 4q_1q_3 - q_2^2 = p_1 a^2 + p_2a + p_3 $$ where \begin{align*} p_1 &= 4h^2(rs-\mu^2),\\ p_2 &= -4\,{h}^{2}{r}^{2}s+4\,{h}^{2}rs\mu+4\,{h}^{2}r{\mu}^{2}-4\,{h}^{2}{\mu}^{3},\\ p_3 &= 4\,{h}^{2}{r}^{2}s{\mu}^{2}-4\,{h}^{2}r{\mu}^{4}-{h}^{2}{r}^{2}{s}^{2}+2\, {h}^{2}rs{\mu}^{2}-{h}^{2}{\mu}^{4}\\ &\qquad -16\,hr{\mu}^{2}+4\,h{r}^{2}+4\,hrs-8\,hr \mu. \end{align*} We hope to find $h$ such that $p_1 > 0$ and $4p_1p_3 - p_2^2 \ge 0$. We have $$4p_1p_3 - p_2^2 = 16(r + s - 2\mu - 4\mu^2)(rs - \mu^2)r h^3[4 - (rs - \mu^2)h].$$ We let $4 - (rs - \mu^2)h = 0$ to get $h = \frac{4}{rs - \mu^2}$ which is what we want.