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Question Number 188535 by Rupesh123 last updated on 03/Mar/23

Answered by mr W last updated on 03/Mar/23

$$duetosymmetry$$$$atextremum$$$$a=b=c=x>0$$$$S=\frac{3{(x+1)}^3}{x}=\frac{3(x+\frac{1}{2}+\frac{1}{2})}{x}$$$$⩾\frac{3{\left(3\sqrt[3]{x\times \frac{1}{2}\times \frac{1}{2}}\right)}^3}{x}$$$$=3\times 27\times \frac{1}{4}=\frac{81}{4}$$$$\Rightarrow minimum=\frac{81}{4}$$

Commented by Rupesh123 last updated on 03/Mar/23

Excellent

Commented by cortano12 last updated on 04/Mar/23

$$\frac{3{(\mathrm{x}+1)}^3}{\mathrm{x}}=\frac{3{(\mathrm{x}+\frac{1}{2}+\frac{1}{2})}^3}{\mathrm{x}}?$$

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