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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
$${a},{b},{c}\:{are}\:{positive}\:{real}\:{numbers} \\ $$$${And} \\ $$$$\frac{\mathrm{1}}{{a}+{b}+\mathrm{1}}+\frac{\mathrm{1}}{{b}+{c}+\mathrm{1}}+\frac{\mathrm{1}}{{a}+{c}+\mathrm{1}}\geqslant\mathrm{1} \\ $$$${prove}\:{that}\:{a}+{b}+{c}\geqslant{ab}+{bc}+{ac} \\ $$
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)
$$\left({a}+{b}+\mathrm{1}\right)\left({a}+{b}+{c}^{\mathrm{2}} \right)\geqslant\left({a}+{b}+{c}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{a}+{b}+\mathrm{1}}\leqslant\frac{{a}+{b}+{c}^{\mathrm{2}} }{\left({a}+{b}+{c}\right)^{\mathrm{2}} } \\ $$$$\left({b}+{c}+\mathrm{1}\right)\left({b}+{c}+{a}^{\mathrm{2}} \right)\geqslant\left({b}+{c}+{a}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{b}+{c}+\mathrm{1}}\leqslant\frac{\left({b}+{c}+{a}^{\mathrm{2}} \right)}{\left({a}+{b}+{c}\right)^{\mathrm{2}} } \\ $$$$\left({a}+{c}+\mathrm{1}\right)\left({a}+{c}+{b}^{\mathrm{2}} \right)\geqslant\left({a}+{c}+{b}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{a}+{c}+\mathrm{1}}\leqslant\frac{\left({a}+{c}+{b}^{\mathrm{2}} \right)}{\left({a}+{b}+{c}\right)^{\mathrm{2}} } \\ $$$$\underset{{cyc}} {\sum}\frac{\mathrm{1}}{{a}+{b}+\mathrm{1}}\leqslant\frac{\mathrm{1}}{\left({a}+{b}+{c}\right)^{\mathrm{2}} }\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({a}+{b}+{c}\right)\right) \\ $$$$\frac{\mathrm{1}}{\left({a}+{b}+{c}\right)^{\mathrm{2}} }\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({a}+{b}+{c}\right)\right)\geqslant\mathrm{1} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\leqslant\mathrm{2}\left({a}+{b}+{c}\right) \\ $$$$\mathrm{2}\left({ab}+{bc}+{ac}\right)\leqslant\mathrm{2}\left({a}+{b}+{c}\right) \\ $$
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
$$\left({a}+{b}+\mathrm{1}\right)\left({ac}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}+{b}^{\mathrm{2}} {c}^{\mathrm{2}} \right)\geqslant\left({ac}+{ab}+{bc}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{a}+{b}+\mathrm{1}}\leqslant\frac{{ac}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}+{b}^{\mathrm{2}} {c}^{\mathrm{2}} }{\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} } \\ $$$$\left({b}+{c}+\mathrm{1}\right)\left({bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)\geqslant\left({bc}+{ac}+{ab}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{b}+{c}+\mathrm{1}}\leqslant\frac{\left({bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)}{\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} } \\ $$$$\left({a}+{c}+\mathrm{1}\right)\left({ab}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}+{a}^{\mathrm{2}} {c}^{\mathrm{2}} \right)\geqslant\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{a}+{c}+\mathrm{1}}\leqslant\frac{\left({ab}^{\mathrm{2}} +{bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} \right)}{\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} } \\ $$$$\underset{{cyc}} {\sum}\frac{\mathrm{1}}{{a}+{b}+\mathrm{1}}\leqslant\frac{\mathrm{1}}{\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} }\left({ac}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}+{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{ab}^{\mathrm{2}} +{bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} \right) \\ $$$$\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ac}\right)={a}^{\mathrm{2}} {b}+{a}^{\mathrm{2}} {c}+{ab}^{\mathrm{2}} +{bc}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}+{ac}^{\mathrm{2}} +\mathrm{3}{abc} \\ $$$${a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} \:¿\:\mathrm{3}{abc} \\ $$$$\underset{{cyc}} {\sum}\frac{\mathrm{1}}{\mathrm{2}}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{b}^{\mathrm{2}} {c}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{b}^{\mathrm{2}} {c}^{\mathrm{2}} \geqslant\mathrm{4}^{\mathrm{4}} \sqrt{\frac{\left({abc}\right)^{\mathrm{4}} }{\mathrm{2}^{\mathrm{4}} }}=\mathrm{2}{abc} \\ $$$$\mathrm{2}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} \right)\:\geqslant\:\mathrm{6}{abc} \\ $$$$\therefore\:{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} \:\geqslant\:\mathrm{3}{abc} \\ $$$$\therefore\:\:\frac{\mathrm{1}}{\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} }\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ac}\right)\geqslant\mathrm{1} \\ $$$${a}+{b}+{c}\geqslant{ab}+{bc}+{ac} \\ $$

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