Let a,b,c ≥ 0 and a+b+c = 1.Prove that (1/(a^2 +1))+(1/(b^2 +1))+(1/(c^2 +1)) ≥ (5/2)


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Question Number 142577 by loveineq last updated on 02/Jun/21

$Let a, b, c ⩾ 0 and a + b + c = 1 . Prove that$ $\frac{1}{a^{2} + 1} + \frac{1}{b^{2} + 1} + \frac{1}{c^{2} + 1} ⩾ \frac{5}{2}$
	Answered by ajfour last updated on 02/Jun/21

	$l . h . s = \frac{1 / a}{a + \frac{1}{a}} + \frac{1 / b}{b + \frac{1}{b}} + \frac{1 / c}{c + \frac{1}{c}}$ $= 3 - (\frac{a}{a + \frac{1}{a}} + \frac{b}{b + \frac{1}{b}} + \frac{c}{c + \frac{1}{c}})$ $⩾ 3 - \frac{1}{2} (a + b + c) ⩾ \frac{5}{2}$ $a s a + \frac{1}{a} ⩾ 2 i f a > 0 .$
	Commented by loveineq last updated on 02/Jun/21

	$But this can not explain that equality holds iff (a, b, c) = (0, 0, 1)$
	Answered by 1549442205PVT last updated on 03/Jun/21
	$i)if (a,b,c)∈A={(0,0,1),(0,1,0),(1,0,0)} we have equality occurs if one of three numbers equal to zero for example c=0 then we need prove for a+b=1:(1/(a^2 +1))+(1/(b^2 +1))≥(3/2) ⇔ 2(a^2 +b^2 +2)≥3(a^2 b^2 +a^2 +b^2 +1)⇔ 3a^2 b^2 +a^2 +b^2 ≤1=(a+b)^2 =a^2 +b^2 +2ab ⇔3a^2 b^2 −2ab≤0⇔ab(3ab−2)≤0 that is true because 3ab≤4ab≤(a+b)^2 =1<2 ii)Consider a,b,c−≠0.Then (1/(a^2 +1))+(1/(b^2 +1))+(1/(c^2 +1)) ≥ (5/2)⇔ −( (1/(a^2 +1))+(1/(b^2 +1))+(1/(c^2 +1))) ≤− (5/2)⇔ 3−( (1/(a^2 +1))+(1/(b^2 +1))+(1/(c^2 +1)))≤3− (5/2)⇔ (a^2 /(a^2 +1))+(b^2 /(b^2 +1))+(c^2 /(c^2 +1))≤(1/2)(2) L.H.S(2)≤(a^2 /(2a))+(b^2 /(2b))+(c^2 /(2c))=((a+b+c)/2)=(1/2) ⇔ (1/(a^2 +1))+(1/(b^2 +1))+(1/(c^2 +1)) ≥ (5/2)(q.e.d) The equality occurs if and if (a,b,c)∈A$
	$i) if (a, b, c) \in A = {(0, 0, 1), (0, 1, 0), (1, 0, 0)}$ $we have equality occurs$ $if one of three numbers equal to zero$ $for example c = 0 then we need prove$ $for a + b = 1 : \frac{1}{a^{2} + 1} + \frac{1}{b^{2} + 1} ⩾ \frac{3}{2} \Leftrightarrow$ $2 (a^{2} + b^{2} + 2) ⩾ 3 (a^{2} b^{2} + a^{2} + b^{2} + 1) \Leftrightarrow$ ${3 a}^{2} b^{2} + a^{2} + b^{2} ⩽ 1 = {(a + b)}^{2} = a^{2} + b^{2} + 2 ab$ $\Leftrightarrow {3 a}^{2} b^{2} - 2 ab ⩽ 0 \Leftrightarrow ab (3 ab - 2) ⩽ 0 that$ $is true because 3 ab ⩽ 4 ab ⩽ {(a + b)}^{2} = 1 < 2$ $ii) Consider a, b, c - \neq 0 . Then$ $\frac{1}{a^{2} + 1} + \frac{1}{b^{2} + 1} + \frac{1}{c^{2} + 1} ⩾ \frac{5}{2} \Leftrightarrow$ $- (\frac{1}{a^{2} + 1} + \frac{1}{b^{2} + 1} + \frac{1}{c^{2} + 1}) ⩽ - \frac{5}{2} \Leftrightarrow$ $3 - (\frac{1}{a^{2} + 1} + \frac{1}{b^{2} + 1} + \frac{1}{c^{2} + 1}) ⩽ 3 - \frac{5}{2} \Leftrightarrow$ $\frac{a^{2}}{a^{2} + 1} + \frac{b^{2}}{b^{2} + 1} + \frac{c^{2}}{c^{2} + 1} ⩽ \frac{1}{2} (2)$ $L . H . S (2) ⩽ \frac{a^{2}}{2 a} + \frac{b^{2}}{2 b} + \frac{c^{2}}{2 c} = \frac{a + b + c}{2} = \frac{1}{2}$ $\Leftrightarrow \frac{1}{a^{2} + 1} + \frac{1}{b^{2} + 1} + \frac{1}{c^{2} + 1} ⩾ \frac{5}{2} (q . e . d)$ $The equality occurs if and if (a, b, c) \in A$
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