Relaxation Problem Number 2

From AM-GM inequality, we get

$a+b+c \geq 3 \sqrt[3]{abc}$

From the question $a+b+c=abc=x$ , we get

$x \geq 3\sqrt[3]{x}$

$x^{3} - 27x \geq 0$

$x(x-3\sqrt{3})(x+3\sqrt{3}) \geq 0$

$x \in [-3\sqrt{3},0] \cup [3\sqrt{3},\infty)$

But $x \geq 0$ from $a,b,c$ are positive reals, the only possible case we get is

$x \geq \boxed{3\sqrt{3}}$ .

It's a well-known fact that $\tan(\alpha)+\tan(\beta)+\tan(\gamma)=\tan(\alpha)\tan(\beta)\tan(\gamma)$ if and only if $\alpha+\beta+\gamma=\pi$ (or, in other words, $\alpha,\beta,\gamma$ are the interior angles of a particular triangle). This beatiful identity suggests the substitution $\alpha=\arctan(a), \beta=\arctan(b)$ and $\gamma=\arctan(c)$ ; so we want to minimize the function $f(\alpha,\beta,\gamma)=\tan(\alpha)+\tan(\beta)+\tan(\gamma)=\tan(\alpha)\tan(\beta)\tan(\gamma)$ under the constraint $\alpha+\beta+\gamma=\pi$ . Recalling that the tangent is convex in $[0,\pi[$ , by Jensen's Inequality (with weights 1/3,1/3,1/3) we have $\frac{\tan(\alpha)}{3}+\frac{\tan(\beta)}{3}+\frac{\tan(\gamma)}{3}\geq \tan\left(\frac{\alpha+\beta+\gamma}{3}\right)$ Which is $f(\alpha,\beta,\gamma)\geq 3\tan\left(\frac{\pi}{3}\right)=3\sqrt{3}$ It follows that $M=3\sqrt{3}=5.1961...$ , so the answer is $\lfloor100M\rfloor=\boxed{519}$

$\text{If we take a=b=c then we have} \\3*\dfrac{x}{3}=(\dfrac{x}{3})^3~~ \implies x^2=27~~~x=3\sqrt{27} =M~~~~~~\boxed{519}~~ \\ \text{ Minimum integer x we know is 6}$ .