Suppose $a,b$ are positive reals.If $3a^2+2b^2=3a+2b$ then find the minimum value of $\sqrt{\frac{a}{b(3a+2)}}+\sqrt{\frac{b}{a(3b+2)}}$

I feel this can be done by AM-GM.I failed to use that $3a^2+2b^2=3a+2b$. Some tricky inequalities are needed here.

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    $\begingroup$ One hint is that $f(a,b) := 3a^2 + 2b^2 - 3a - 2b$ is a paraboloid of revolution. So you a trying to maximize the function with the square roots over this paraboloid. $\endgroup$ – Viktor Glombik Apr 14 at 13:48
  • $\begingroup$ @ViktorGlombik Do anyone have Basic solution without calculus $\endgroup$ – Sufaid Saleel Apr 14 at 14:01

Just an ugly exercise for numerical calculation (I gave up with Lagrange multipliers).

Use the constraint to eliminate $b$ as a function of $a$ $$b_\pm=\frac{1}{2} \left(1\pm\sqrt{-6 a^2+6 a+1}\right)$$ Using $b=b_+$, you are left with $$f(a)= \sqrt{\frac{2a}{(3 a+2) \left(1+\sqrt{-6 a^2+6 a+1}\right)}}+\sqrt{\frac{1+\sqrt{-6 a^2+6 a+1}}{7a+3a \sqrt{-6 a^2+6 a+1} }}$$

Computing $f'(a)$ (a small nightmare) and looking for its zero, starting with $a_0=1$, Newton iterates are $$\left( \begin{array}{cc} n & a_n \\ 0 & 1.0000000000 \\ 1 & 0.9657338664 \\ 2 & 0.9591954631 \\ 3 & 0.9590355426 \\ 4 & 0.9590354535 \end{array} \right)$$ leading to $b= 1.0558144283$ and the minimum value already reported by Michael Rozenberg.


Assuming that we know that the solution is close to $(1,1)$, let $a=1+\alpha$ and $b=1+\beta$. Expand as Taylor series the constraint to get $$3 \alpha +2 \beta=0\implies \beta=-\frac 32 \alpha$$ This makes the function to be $$g(\alpha)= \sqrt{\frac{2(\alpha +1)}{-9 \alpha ^2-9 \alpha +10}}+\sqrt{\frac{2-3 \alpha }{-9 \alpha ^2+\alpha +10}}$$ Using Taylor expansion $$g(\alpha)=\frac{2}{\sqrt{5}}+\frac{3 \alpha }{20 \sqrt{5}}+\frac{851 \alpha ^2}{800 \sqrt{5}}+O\left(\alpha ^3\right)$$ $$g'(\alpha)=\frac{3}{20 \sqrt{5}}+\frac{851 \alpha }{400 \sqrt{5}}+O\left(\alpha ^2\right) \implies \alpha=-\frac{60}{851}\qquad \beta=\frac{90}{851}$$ That is to say $a=\frac{791}{851}\approx 0.929495$ and $b=\frac{941}{851}\approx 1.10576$.

Plugging in $g(\alpha)$ leads to $$g\left(-\frac{60}{851}\right)=\frac{1}{5} \left(\sqrt{\frac{673141}{153383}}+\sqrt{\frac{800791}{143171}}\right)\approx 0.891982$$


Without using the expansion for the constraint, using $b=b_+$ and performing around $a=1$ a Taylor expansion of $f(a)$, we obtain $$f(a)=\frac{2}{\sqrt{5}}+\frac{3 (a-1)}{20 \sqrt{5}}+\frac{1751 (a-1)^2}{800 \sqrt{5}}+\frac{23623 (a-1)^3}{3200 \sqrt{5}}+\frac{9170799 (a-1)^4}{256000 \sqrt{5}}+O\left((a-1)^5\right)$$ which is minimum for $$a\approx 0.959646 \implies b\approx 1.05506$$ and a minimum value equal to $0.893133$.

The exact solution is $a\approx 0.959035$, $b\approx 1.05581$ and $f_{min}\approx 0.893132$.

  • $\begingroup$ Can anyone do it WITHOUT using CALCULUS $\endgroup$ – Sufaid Saleel Apr 14 at 14:38
  • $\begingroup$ @SufaidSaleel. Not me ! Cheers. $\endgroup$ – Claude Leibovici Apr 14 at 14:39
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    $\begingroup$ @Sufaid Saleel At least, now do you see that the equality does not occurs for $a= b$? Thank you, Claude! +1 $\endgroup$ – Michael Rozenberg Apr 14 at 15:51

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