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Question Number 193965 by Subhi last updated on 24/Jun/23
a,b,c are positive real numbers And (1/(a+b+1))+(1/(b+c+1))+(1/(a+c+1))≥1 prove that a+b+c≥ab+bc+ac
a,b,carepositiverealnumbersAnd1a+b+1+1b+c+1+1a+c+1⩾1provethata+b+c⩾ab+bc+ac
Answered by Subhi last updated on 26/Jun/23
(a+b+1)(a+b+c^2 )≥(a+b+c)^2 (1/(a+b+1))≤((a+b+c^2 )/((a+b+c)^2 )) (b+c+1)(b+c+a^2 )≥(b+c+a)^2 (1/(b+c+1))≤(((b+c+a^2 ))/((a+b+c)^2 )) (a+c+1)(a+c+b^2 )≥(a+c+b)^2 (1/(a+c+1))≤(((a+c+b^2 ))/((a+b+c)^2 )) Σ_(cyc) (1/(a+b+1))≤(1/((a+b+c)^2 ))(a^2 +b^2 +c^2 +2(a+b+c)) (1/((a+b+c)^2 ))(a^2 +b^2 +c^2 +2(a+b+c))≥1 (a+b+c)^2 −(a^2 +b^2 +c^2 )≤2(a+b+c) 2(ab+bc+ac)≤2(a+b+c)
(a+b+1)(a+b+c2)⩾(a+b+c)21a+b+1⩽a+b+c2(a+b+c)2(b+c+1)(b+c+a2)⩾(b+c+a)21b+c+1⩽(b+c+a2)(a+b+c)2(a+c+1)(a+c+b2)⩾(a+c+b)21a+c+1⩽(a+c+b2)(a+b+c)2∑cyc1a+b+1⩽1(a+b+c)2(a2+b2+c2+2(a+b+c))1(a+b+c)2(a2+b2+c2+2(a+b+c))⩾1(a+b+c)2−(a2+b2+c2)⩽2(a+b+c)2(ab+bc+ac)⩽2(a+b+c)
Answered by Subhi last updated on 26/Jun/23
(a+b+1)(ac^2 +a^2 b+b^2 c^2 )≥(ac+ab+bc)^2 (1/(a+b+1))≤((ac^2 +a^2 b+b^2 c^2 )/((ab+bc+ac)^2 )) (b+c+1)(bc^2 +a^2 c+a^2 b^2 )≥(bc+ac+ab)^2 (1/(b+c+1))≤(((bc^2 +a^2 c+a^2 b^2 ))/((ab+bc+ac)^2 )) (a+c+1)(ab^2 +b^2 c+a^2 c^2 )≥(ab+bc+ac)^2 (1/(a+c+1))≤(((ab^2 +bc^2 +a^2 c^2 ))/((ab+bc+ac)^2 )) Σ_(cyc) (1/(a+b+1))≤(1/((ab+bc+ac)^2 ))(ac^2 +a^2 b+b^2 c^2 +bc^2 +a^2 c+a^2 b^2 +ab^2 +bc^2 +a^2 c^2 ) (a+b+c)(ab+bc+ac)=a^2 b+a^2 c+ab^2 +bc^2 +b^2 c+ac^2 +3abc a^2 b^2 +b^2 c^2 +a^2 c^2 ¿ 3abc Σ_(cyc) (1/2)a^2 b^2 +(1/2)a^2 b^2 +(1/2)b^2 c^2 +(1/2)b^2 c^2 ≥4^4 (√(((abc)^4 )/2^4 ))=2abc 2(a^2 b^2 +b^2 c^2 +a^2 c^2 ) ≥ 6abc ∴ a^2 b^2 +b^2 c^2 +a^2 c^2 ≥ 3abc ∴ (1/((ab+bc+ac)^2 ))(a+b+c)(ab+bc+ac)≥1 a+b+c≥ab+bc+ac
(a+b+1)(ac2+a2b+b2c2)⩾(ac+ab+bc)21a+b+1⩽ac2+a2b+b2c2(ab+bc+ac)2(b+c+1)(bc2+a2c+a2b2)⩾(bc+ac+ab)21b+c+1⩽(bc2+a2c+a2b2)(ab+bc+ac)2(a+c+1)(ab2+b2c+a2c2)⩾(ab+bc+ac)21a+c+1⩽(ab2+bc2+a2c2)(ab+bc+ac)2∑cyc1a+b+1⩽1(ab+bc+ac)2(ac2+a2b+b2c2+bc2+a2c+a2b2+ab2+bc2+a2c2)(a+b+c)(ab+bc+ac)=a2b+a2c+ab2+bc2+b2c+ac2+3abca2b2+b2c2+a2c2¿3abc∑cyc12a2b2+12a2b2+12b2c2+12b2c2⩾44(abc)424=2abc2(a2b2+b2c2+a2c2)⩾6abc∴a2b2+b2c2+a2c2⩾3abc∴1(ab+bc+ac)2(a+b+c)(ab+bc+ac)⩾1a+b+c⩾ab+bc+ac

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