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Question Number 144603 by loveineq last updated on 26/Jun/21

Let a,b,c > 0 and (a+b)(b+c) = 4. Prove that (2a+b)(a+b)+(b+2c)(b+c) ≥ 8+(1/2)(a+2b+c)(c+a) Determine when equality holds.

$$\mathrm{Let}\:{a},{b},{c}\:>\:\mathrm{0}\:\mathrm{and}\:\left({a}+{b}\right)\left({b}+{c}\right)\:=\:\mathrm{4}.\:\mathrm{Prove}\:\mathrm{that} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}{a}+{b}\right)\left({a}+{b}\right)+\left({b}+\mathrm{2}{c}\right)\left({b}+{c}\right)\:\geqslant\:\mathrm{8}+\frac{\mathrm{1}}{\mathrm{2}}\left({a}+\mathrm{2}{b}+{c}\right)\left({c}+{a}\right)\:\:\:\:\:\:\: \\ $$ $$\mathrm{Determine}\:\mathrm{when}\:\mathrm{equality}\:\mathrm{holds}.\:\:\: \\ $$

Answered by mindispower last updated on 27/Jun/21

(2a+b)(a+b)+(b+2c)(b+c) =(a+b)^2 +(b+c)^2 +a^2 +ab+bc+c^2 (a+b)^2 +(b+c)^2 ≥2(a+b)(b+c)=8 a^2 +c^2 +ab+bc=b(a+c)+a^2 +c^2 ..E a^2 +c^2 ≥(((a+c)^2 )/2) E⇔a^2 +c^2 +ab+bc≥(((a+c)^2 )/2)+b(a+c)=(a+c)(b+((a+c)/2)) =(1/2)(a+c)(2b+a+c) ⇒≥(1/2)(a+c)(2b+a+c)+8

$$\left(\mathrm{2}{a}+{b}\right)\left({a}+{b}\right)+\left({b}+\mathrm{2}{c}\right)\left({b}+{c}\right) \\ $$ $$=\left({a}+{b}\right)^{\mathrm{2}} +\left({b}+{c}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} +{ab}+{bc}+{c}^{\mathrm{2}} \\ $$ $$\left({a}+{b}\right)^{\mathrm{2}} +\left({b}+{c}\right)^{\mathrm{2}} \geqslant\mathrm{2}\left({a}+{b}\right)\left({b}+{c}\right)=\mathrm{8} \\ $$ $${a}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ab}+{bc}={b}\left({a}+{c}\right)+{a}^{\mathrm{2}} +{c}^{\mathrm{2}} ..{E} \\ $$ $${a}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant\frac{\left({a}+{c}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$ $${E}\Leftrightarrow{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ab}+{bc}\geqslant\frac{\left({a}+{c}\right)^{\mathrm{2}} }{\mathrm{2}}+{b}\left({a}+{c}\right)=\left({a}+{c}\right)\left({b}+\frac{{a}+{c}}{\mathrm{2}}\right) \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{c}\right)\left(\mathrm{2}{b}+{a}+{c}\right) \\ $$ $$\Rightarrow\geqslant\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{c}\right)\left(\mathrm{2}{b}+{a}+{c}\right)+\mathrm{8} \\ $$ $$ \\ $$ $$ \\ $$

Commented byloveineq last updated on 27/Jun/21

thanks

$${thanks} \\ $$

Commented bymindispower last updated on 27/Jun/21

pleasur

$${pleasur} \\ $$

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