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Question Number 144634 by loveineq last updated on 27/Jun/21
Let a,b,c > 0 and (a+b)(b+c) = 4. Prove that (1) a^2 +2b^2 +c^2 +((2b(c^2 +a^2 ))/(c+a)) ≥ 6 (2) a^3 +3b^3 +c^3 +((3b(c^3 +a^3 ))/(c+a)) ≥ 8 (3) a^4 +4b^4 +c^4 +((4b(c^4 +a^4 ))/(c+a)) ≥ 10
$$\mathrm{Let}\:{a},{b},{c}\:>\:\mathrm{0}\:\mathrm{and}\:\left({a}+{b}\right)\left({b}+{c}\right)\:=\:\mathrm{4}.\:\mathrm{Prove}\:\mathrm{that}\:\:\:\:\:\:\:\: \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\frac{\mathrm{2}{b}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}{{c}+{a}}\:\geqslant\:\mathrm{6} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{3}} +\mathrm{3}{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +\frac{\mathrm{3}{b}\left({c}^{\mathrm{3}} +{a}^{\mathrm{3}} \right)}{{c}+{a}}\:\geqslant\:\mathrm{8} \\ $$$$\left(\mathrm{3}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{4}} +\mathrm{4}{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +\frac{\mathrm{4}{b}\left({c}^{\mathrm{4}} +{a}^{\mathrm{4}} \right)}{{c}+{a}}\:\geqslant\:\mathrm{10} \\ $$
Answered by mindispower last updated on 27/Jun/21
c^2 +a^2 ≥(1/2)(c+a)^2 (1)⇔≥a^2 +2b^2 +c^2 +b(c+a) a^2 +c^2 ≥ac+(1/2)(a^2 +c^2 ) ≥(b^2 +bc+ac+cb)+b^2 +(1/2)(a^2 +c^2 ) b^2 +bc+ac+cb+((a^2 +c^2 )/2)=t (b+a)(b+c)+(1/4)(b^2 +a^2 )+(1/4)(b^2 +c^2 )+(1/2)b^2 +(1/4)(a^2 +c^2 ) ≥4+(1/4).2ab+(1/4).2bc+(1/2)b^2 +((2ac)/4)=4+(1/2)(b^2 +ac+bc+ab) =4+(1/2)(b+a)(b+c)=6 ⇔≥6
$${c}^{\mathrm{2}} +{a}^{\mathrm{2}} \geqslant\frac{\mathrm{1}}{\mathrm{2}}\left({c}+{a}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\geqslant{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{b}\left({c}+{a}\right) \\ $$$${a}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant{ac}+\frac{\mathrm{1}}{\mathrm{2}}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$\geqslant\left({b}^{\mathrm{2}} +{bc}+{ac}+{cb}\right)+{b}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$${b}^{\mathrm{2}} +{bc}+{ac}+{cb}+\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}}={t} \\ $$$$\left({b}+{a}\right)\left({b}+{c}\right)+\frac{\mathrm{1}}{\mathrm{4}}\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{4}}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}{b}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$\geqslant\mathrm{4}+\frac{\mathrm{1}}{\mathrm{4}}.\mathrm{2}{ab}+\frac{\mathrm{1}}{\mathrm{4}}.\mathrm{2}{bc}+\frac{\mathrm{1}}{\mathrm{2}}{b}^{\mathrm{2}} +\frac{\mathrm{2}{ac}}{\mathrm{4}}=\mathrm{4}+\frac{\mathrm{1}}{\mathrm{2}}\left({b}^{\mathrm{2}} +{ac}+{bc}+{ab}\right) \\ $$$$=\mathrm{4}+\frac{\mathrm{1}}{\mathrm{2}}\left({b}+{a}\right)\left({b}+{c}\right)=\mathrm{6} \\ $$$$\Leftrightarrow\geqslant\mathrm{6} \\ $$$$ \\ $$$$ \\ $$
Answered by mindispower last updated on 27/Jun/21
(2) i think ((3b(c^3 +a^3 ))/(c+a))....? (3)...(4) Weighted AM worcks
$$\left(\mathrm{2}\right)\:{i}\:{think}\:\frac{\mathrm{3}{b}\left({c}^{\mathrm{3}} +{a}^{\mathrm{3}} \right)}{{c}+{a}}….? \\ $$$$\left(\mathrm{3}\right)…\left(\mathrm{4}\right)\:{Weighted}\:{AM}\:{worcks} \\ $$$$ \\ $$
Commented by loveineq last updated on 27/Jun/21
yes, is ((c^3 +a^3 )/(c+a)).
$${yes},\:{is}\:\frac{{c}^{\mathrm{3}} +{a}^{\mathrm{3}} }{{c}+{a}}. \\ $$

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