Geometry in inequality

Since $a,b,c > 0$ , we can find $0 < A,B,C < \tfrac12\pi$ such that $a = \tan A$ , $b = \tan B$ and $c = \tan C$ . But since $\tan(A+B+C) \; = \; \frac{a+b+c - abc}{1 - ab - ac - bc} \; = \; 0 \;,$ it follows that $A+B+C = \pi$ , so that $A,B,C$ are the angles of an acute-angled triangle. Thus the question asks us to maximize $2\cos A + \cos B + \cos C$ over all such triangles. We have $\begin{array}{rcl} 2\cos A + \cos B + \cos C & = & 2\cos\tfrac{B+C}{2} \cos\tfrac{B-C}{2} + 2\cos A \\ & \le & 2\cos\tfrac{B+C}2 + 2\cos A \\ & = & 2\cos\tfrac{B+C}2 + 2\cos(\pi - B-C) \\ & = & 2\cos\tfrac{B+C}2 - 2\cos(B+C) \\ & = & 2\cos\tfrac{B+C}2 - 4\cos^2\tfrac{B+C}2 + 2 \\ & = & \tfrac94 - \left(2\cos\tfrac{B+C}2 - \tfrac12\right)^2 \end{array}$ so we deduce that $2\cos A + \cos B + \cos C \,\le \, \tfrac94$ , with equality when $B=C$ and $\cos\tfrac{B+C}2 = \tfrac14$ . Thus the correct value of $P$ is $\tfrac94$ , and so $10000P \,=\, 22500$ . There is no need for the integer part.

$\small{\color{goldenrod}{\text{Let a=tanA,b=tanB and c=tanC s.t 0 }\leq A,B,C \leq \dfrac{\pi}{2}}}$ a+b+c=abc simply states that A,B,C are angles of a triangle and we have to find maximum of 2tanA+tanB+tanC. Using $\color{forestgreen}{\text{Using Method of Lagrange's Multipliers}}$

2tanA+tanB+tanC= $\lambda$ tanAtanBtanC, under the constraint that they obey Angle sum property i.e A+B+C= $\pi$

Solving we get, $2\sin A=\sin B=\sin C=-\lambda$ $\Rightarrow B=C ~and ~\sin B=2\sin A=2 \sin(\pi-2B)$ $\Rightarrow$ cosB=cosC= $\dfrac{1}{4}$ and cosA= $\dfrac{7}{8}.$ Hence maximum value occurs when $\small{\color{#3D99F6}{\cos B=\cos C=\dfrac{1}{4} and \cos A=\dfrac{7}{8}}}$ Hence, Maximum value= $2(\dfrac{7}{8})+\dfrac{1}{4}+\dfrac{1}{4}=2.25$ $\Large{10000\times 2.25=22500}$