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Question Number 92835 by unknown last updated on 09/May/20

Commented by unknown last updated on 09/May/20

$$\mathrm{let}a,b,c\mathrm{is}\mathrm{real}\mathrm{positive}\mathrm{number}.a+b+c=3$$$$\mathrm{find}\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}:$$$$A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}$$$$$$

Commented by mr W last updated on 10/May/20

$$a,b,c>0$$$$abc⩽{\left(\frac{a+b+c}{3}\right)}^3={\left(\frac{3}{3}\right)}^3=1$$$$A=\frac{2(ab+bc+ca)}{abc}-(a^2+b^2+c^2)$$$$A=\frac{{(a+b+c)}^2-(a^2+b^2+c^2)}{abc}-(a^2+b^2+c^2)$$$$A⩾9-2(a^2+b^2+c^2)⩾9-6{\left(abc\right)}^{\frac{2}{3}}⩾9-6=3$$

Commented by unknown last updated on 11/May/20

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{sir}$$

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