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

   ⇃

$$\:\:\:\downharpoonleft\underline{\:} \\ $$

Answered by A5T last updated on 19/Dec/24

(√(a^2 +b^2 ))+(√(c^2 +d^2 ))≥(√((a+c)^2 +(b+d)^2 ))  ⇒(√(2^2 +x^2 ))+(√(3^2 +y^2 ))≥(√((2+3)^2 +(x+y)^2 ))=13

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

Commented by efronzo1 last updated on 20/Dec/24

by Euclid theorem?

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

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

∣AB∣+∣BC∣≥∣AC∣

$$\mid{AB}\mid+\mid{BC}\mid\geqslant\mid{AC}\mid \\ $$

Answered by golsendro last updated on 20/Dec/24

  f(x)=(√(x^2 +2^2 )) + (√((12−x)^2 +3^2 ))    min when (x/2) = ((12−x)/3) ⇒x=((24)/5)    min value =(√((((24)/5))^2 +4)) +(√((12−((24)/5))^2 +9))   = 13

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

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