Find the minimum value of $$f(x) = {3\over \sqrt{x}+1} - {12\over \sqrt{x}+3}$$

The domain of $f(x)$ is $x ∈ (0,∞)$. Then, using derivatives, I can find the minimum value is $f(1)=-1.5$. However, this uses derivatives.


If $t=\sqrt{x}\geq 0$ we get $$f(x) = {3\over \sqrt{x}+1} - {12\over \sqrt{x}+3}$$

$$ = {3\over t+1} - {12\over t+3} $$

$$ = {3(t+3)-12(t+1)\over (t+1)(t+3)} $$

$$ = {-3(3t+1)\over (t+1)(t+3)} $$

Now try to find such real $m$ that quadratic equation $${-3(3t+1)\over (t+1)(t+3)} =m$$ i.e. $$mt^2+(4m+9)t +3m+3=0$$ has exactly one solution, so the discriminant $$4m^2+60m+81=0$$

and we get $m_1 = -{3\over 2}$ (at $t=1$ ) and $m_2 = -{27\over 2}$ (at $t =-{27\over 2}<0$) so this on can not be. So the minimum value of this expression is $\boxed{-{3\over 2}}$


Hint: Start with $$(\sqrt{x}-1)^2\geq 0$$ then we get $$3+3x\geq 6\sqrt{x}$$ and then $$18\sqrt{x}+27+3x\geq 24x+24$$


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