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Question Number 155547 by mathdanisur last updated on 02/Oct/21

let a;b;c>0 and a+b+c=3 find min value of the expression: S = abc + (a-1)^2 + (b-1)^2 + (c-1)^2

$$\mathrm{let}\:\:\mathrm{a};\mathrm{b};\mathrm{c}>\mathrm{0}\:\:\mathrm{and}\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{3} \\ $$ $$\mathrm{find}\:\mathrm{min}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expression}: \\ $$ $$\mathrm{S}\:=\:\mathrm{abc}\:+\:\left(\mathrm{a}-\mathrm{1}\right)^{\mathrm{2}} \:+\:\left(\mathrm{b}-\mathrm{1}\right)^{\mathrm{2}} \:+\:\left(\mathrm{c}-\mathrm{1}\right)^{\mathrm{2}} \\ $$

Answered by ghimisi last updated on 02/Oct/21

p=3 p^3 ≥27r⇒r≤1 schur⇒p^3 +9r≥4pq⇒3r−4q≥−9 S=r+a^2 +b^2 +c^2 −2(a+b+c)+3 S=r+p^2 −2q−2p+3=6+r−2q 2S=12+3r−4q−r≥12−9−1=2 ⇒minS=1 (a=b=c=1)

$${p}=\mathrm{3} \\ $$ $${p}^{\mathrm{3}} \geqslant\mathrm{27}{r}\Rightarrow{r}\leqslant\mathrm{1} \\ $$ $${schur}\Rightarrow{p}^{\mathrm{3}} +\mathrm{9}{r}\geqslant\mathrm{4}{pq}\Rightarrow\mathrm{3}{r}−\mathrm{4}{q}\geqslant−\mathrm{9} \\ $$ $${S}={r}+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}\left({a}+{b}+{c}\right)+\mathrm{3} \\ $$ $${S}={r}+{p}^{\mathrm{2}} −\mathrm{2}{q}−\mathrm{2}{p}+\mathrm{3}=\mathrm{6}+{r}−\mathrm{2}{q} \\ $$ $$\mathrm{2}{S}=\mathrm{12}+\mathrm{3}{r}−\mathrm{4}{q}−{r}\geqslant\mathrm{12}−\mathrm{9}−\mathrm{1}=\mathrm{2} \\ $$ $$\Rightarrow{minS}=\mathrm{1}\:\:\:\left({a}={b}={c}=\mathrm{1}\right) \\ $$

Commented bymathdanisur last updated on 02/Oct/21

Very nice Ser, thank you so much

$$\mathrm{Very}\:\mathrm{nice}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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