a + b + c = 1 a, b, c ≤ 1 Prove that (1/(a^2 + 1)) + (1/(b^2 + 1)) + (1/(c^2 + 1)) ≤ ((27)/(10))


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Question Number 58964 by naka3546 last updated on 02/May/19

$a + b + c = 1$ $a, b, c ⩽ 1$ $P r o v e t h a t$ $\frac{1}{a^{2} + 1} + \frac{1}{b^{2} + 1} + \frac{1}{c^{2} + 1} ⩽ \frac{27}{10}$
	Answered by tanmay last updated on 02/May/19

	$\frac{27}{10} = \frac{30 - 3}{10} = (1 + 1 + 1) - (\frac{1}{10} + \frac{1}{10} + \frac{1}{10})$ $w e h a v e t o p r o v e$ $\frac{1}{a^{2} + 1} - 1 + \frac{1}{b^{2} + 1} - 1_{} + \frac{1}{c^{2} + 1} - 1 + \frac{3}{10} < 0$ $\frac{- a^{2}}{a^{2} + 1} + \frac{- b^{2}}{b^{2} + 1} + \frac{- c^{2}}{c^{2} + 1} + \frac{3}{10}$ ${(a - 1)}^{2} > 0$ $a^{2} + 1 > 2 a$ $\frac{1}{a^{2} + 1} < \frac{1}{2 a}$ $\frac{a^{2}}{a^{2} + 1} < \frac{a^{2}}{2 a}$ $- \frac{a^{2}}{a^{2} + 1} > - \frac{a}{2}$ $- (\frac{a^{2}}{a^{2} + 1} + \frac{b^{2}}{b^{2} + 1} + \frac{c^{2}}{c^{2} + 1}) > - (\frac{a}{2} + \frac{b}{2} + \frac{c}{2})$ $- (\frac{a^{2}}{a^{2} + 1} + \frac{b^{2}}{b^{2} + 1} + \frac{c^{2}}{c^{2} + 1}) + \frac{3}{10} > \frac{- 1}{2} + \frac{3}{10}$ $- (\frac{a^{2}}{a^{2} + 1} + \frac{b^{2}}{b^{2} + 1} + \frac{c^{2}}{c^{2} + 1}) + \frac{3}{10} > \frac{- 2}{10}$ $r i g h t h a n d s i d e i s = \frac{- 1}{5} < 0$ $s o \frac{1}{a^{2} + 1} + \frac{1}{b^{2} + 1} + \frac{1}{c^{2} + 1} ⩽ \frac{27}{10}$ $o t h e r s a r e r e q u e s t e d f o r k i n d p e r u s a l p l e a s e$ $a n d c h e c k p l s$
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