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Question Number 214784 by efronzo1 last updated on 19/Dec/24

$$\downharpoonleft \underset{―}{}$$

Answered by A5T last updated on 19/Dec/24

$$\sqrt{a^2+b^2}+\sqrt{c^2+d^2}⩾\sqrt{{(a+c)}^2+{(b+d)}^2}$$$$\Rightarrow \sqrt{2^2+x^2}+\sqrt{3^2+y^2}⩾\sqrt{{(2+3)}^2+{(x+y)}^2}=13$$

Commented by efronzo1 last updated on 20/Dec/24

$$\mathrm{by}\mathrm{Euclid}\mathrm{theorem}?$$

Commented by mr W last updated on 20/Dec/24

$$\mid AB\mid +\mid BC\mid ⩾\mid AC\mid$$

Answered by golsendro last updated on 20/Dec/24

$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{{\mathrm{x}}^2+2^2}+\sqrt{{(12-\mathrm{x})}^2+3^2}$$$$\min \mathrm{when}\frac{\mathrm{x}}{2}=\frac{12-\mathrm{x}}{3}\Rightarrow \mathrm{x}=\frac{24}{5}$$$$\min \mathrm{value}=\sqrt{{\left(\frac{24}{5}\right)}^2+4}+\sqrt{{(12-\frac{24}{5})}^2+9}$$$$=13$$

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