The maximum value of $P = \frac{2}{a^{2} + 1} - \frac{2}{b^{2} + 1} + \frac{3}{c^{2} + 1}$ is _____ where a, b, c > 0 and $a b c + a + c = b$

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$\frac{8}{3}$

$\frac{10}{3}$

answer is D.

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Detailed Solution

$a = \tan A; b = \tan B; c = \tan C; 0 < A, B, C < \frac{π}{2} 0 < A + C < \frac{π}{2} P = 2 \cos^{2} A + 3 \cos^{2} C - 2 \cos^{2} (A + C)$

= $1 + \cos 2 A + \frac{3}{2} + \frac{3}{2} \cos 2 + \cos (2 A + 2 C)$
= $1 + \cos 2 A + \frac{3}{2} + \frac{3}{2} \cos 2 C - \cos 2 A \cos 2 C + \sin 2 A \sin 2 C$
= $\frac{3}{2} + \frac{3}{2} \cos 2 C + (1 - \cos 2 C) (\cos 2 A + \sin 2 C . \sin 2 A)$
$< \frac{3}{2} + \frac{3}{2} \cos 2 C + \sqrt{{(1 - \cos 2 C)}^{2} + {(\sin 2 C)}^{2}}$
= $\frac{3}{2} + \frac{3}{2} \cos 2 C + \sqrt{2 - 2 \cos 2 C} = \frac{5}{2} + \frac{3}{2} \cos 2 C + 2 \sin C$
$\frac{3}{2} + \frac{3}{2} (1 - 2 \sin^{2} C) + 2 \sin C$ , $\frac{3}{2} + \frac{3}{2} - 3 \sin^{2} C + 2 \sin C$
$\frac{3}{2} + \frac{3}{2} - 3 [\sin^{2} C \frac{2}{3} \sin C + \frac{1}{9} - \frac{1}{9}] = + \frac{1}{3} - 3 {[\sec \frac{1}{3}]}^{2}$ = $\frac{10}{3}$