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Question Number 193965 by Subhi last updated on 24/Jun/23

$$a,b,carepositiverealnumbers$$$$And$$$$\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{a+c+1}⩾1$$$$provethata+b+c⩾ab+bc+ac$$

Answered by Subhi last updated on 26/Jun/23

$$(a+b+1)(a+b+c^2)⩾{(a+b+c)}^2$$$$\frac{1}{a+b+1}⩽\frac{a+b+c^2}{{(a+b+c)}^2}$$$$(b+c+1)(b+c+a^2)⩾{(b+c+a)}^2$$$$\frac{1}{b+c+1}⩽\frac{(b+c+a^2)}{{(a+b+c)}^2}$$$$(a+c+1)(a+c+b^2)⩾{(a+c+b)}^2$$$$\frac{1}{a+c+1}⩽\frac{(a+c+b^2)}{{(a+b+c)}^2}$$$$\sum _{cyc}\frac{1}{a+b+1}⩽\frac{1}{{(a+b+c)}^2}(a^2+b^2+c^2+2(a+b+c))$$$$\frac{1}{{(a+b+c)}^2}(a^2+b^2+c^2+2(a+b+c))⩾1$$$${(a+b+c)}^2-(a^2+b^2+c^2)⩽2(a+b+c)$$$$2(ab+bc+ac)⩽2(a+b+c)$$

Answered by Subhi last updated on 26/Jun/23

$$(a+b+1)({ac}^2+a^{2}b+b^2{c}^2)⩾{(ac+ab+bc)}^2$$$$\frac{1}{a+b+1}⩽\frac{{ac}^2+a^{2}b+b^2{c}^2}{{(ab+bc+ac)}^2}$$$$(b+c+1)({bc}^2+a^{2}c+a^2{b}^2)⩾{(bc+ac+ab)}^2$$$$\frac{1}{b+c+1}⩽\frac{({bc}^2+a^{2}c+a^2{b}^2)}{{(ab+bc+ac)}^2}$$$$(a+c+1)({ab}^2+b^{2}c+a^2{c}^2)⩾{(ab+bc+ac)}^2$$$$\frac{1}{a+c+1}⩽\frac{({ab}^2+{bc}^2+a^2{c}^2)}{{(ab+bc+ac)}^2}$$$$\sum _{cyc}\frac{1}{a+b+1}⩽\frac{1}{{(ab+bc+ac)}^2}({ac}^2+a^{2}b+b^2{c}^2+{bc}^2+a^{2}c+a^2{b}^2+{ab}^2+{bc}^2+a^2{c}^2)$$$$(a+b+c)(ab+bc+ac)=a^{2}b+a^{2}c+{ab}^2+{bc}^2+b^{2}c+{ac}^2+3abc$$$$a^2{b}^2+b^2{c}^2+a^2{c}^2¿3abc$$$$\sum _{cyc}\frac{1}{2}{a}^2{b}^2+\frac{1}{2}{a}^2{b}^2+\frac{1}{2}{b}^2{c}^2+\frac{1}{2}{b}^2{c}^2⩾4^4\sqrt{\frac{{\left(abc\right)}^4}{2^4}}=2abc$$$$2(a^2{b}^2+b^2{c}^2+a^2{c}^2)⩾6abc$$$$\therefore a^2{b}^2+b^2{c}^2+a^2{c}^2⩾3abc$$$$\therefore \frac{1}{{(ab+bc+ac)}^2}(a+b+c)(ab+bc+ac)⩾1$$$$a+b+c⩾ab+bc+ac$$

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