Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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 byloveineq last updated on 27/Jun/21

yes, is ((c^3 +a^3 )/(c+a)).

$${yes},\:{is}\:\frac{{c}^{\mathrm{3}} +{a}^{\mathrm{3}} }{{c}+{a}}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com