Question Number 208823 by Ismoiljon_008 last updated on 23/Jun/24
$$ \\ $$$$\:\:\:{If}\:{a}+{b}+{c}=\mathrm{15},\:{then}\:{find}\:{the}\:{smallest}\:{value}\: \\ $$$$\:\:\:{of}\:{the}\:{expression}\:\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}+\sqrt{{b}^{\mathrm{2}} +\mathrm{9}}+\sqrt{{c}^{\mathrm{2}} +\mathrm{16}}. \\ $$$$\:\:\:\:\:{Help}\:{please} \\ $$
Answered by mr W last updated on 23/Jun/24
$$\sqrt{{a}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }+\sqrt{{b}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }+\sqrt{{c}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} } \\ $$$$\geqslant\sqrt{\left({a}+{b}+{c}\right)^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{3}+\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\mathrm{15}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} } \\ $$$$=\mathrm{17}\:={minimum} \\ $$
Commented by Ismoiljon_008 last updated on 24/Jun/24
$${thanks} \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 24/Jun/24
$$\mid\overset{\rightarrow} {{A}}\mid+\mid\overset{\rightarrow} {{B}}\mid+\mid\overset{\rightarrow} {{C}}\mid\geqslant\mid\overset{\rightarrow} {{A}}+\overset{\rightarrow} {{B}}+\overset{\rightarrow} {{C}}\mid \\ $$$${with}\:\overset{\rightarrow} {{A}}=\left({a},\:\mathrm{1}\right),\:\overset{\rightarrow} {{B}}=\left({b},\:\mathrm{3}\right),\:\overset{\rightarrow} {{C}}=\left({c},\:\mathrm{4}\right) \\ $$
Commented by mr W last updated on 24/Jun/24
Answered by A5T last updated on 24/Jun/24
![[E]^2 =[(√(a^2 +1))+(√(b^2 +9))+(√(c^2 +16 ))]^2 =a^2 +b^2 +c^2 +1+9+16+2(√(a^2 +1))(√(b^2 +9))+ 2(√(b^2 +9))(√(c^2 +16))+2(√(c^2 +16))(√(a^2 +1)) ≥a^2 +b^2 +c^2 +26+2(ab+1×3)+2(bc+3×4)+ 2(ac+4×1) =(a+b+c)^2 −2ab−2bc−2ca+2ab+2ac+2bc+26 +6+24+8=15^2 +8^2 =17^2 E^2 ≥17^2 ⇒E≥17 Equality when (a,b,c)=(λ,3λ,4λ)⇒8λ=15 ⇒λ=((15)/8)⇒(a,b,c)=(((15)/8),((45)/8),((15)/2))](https://www.tinkutara.com/question/Q208826.png)
$$\left[{E}\right]^{\mathrm{2}} =\left[\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}+\sqrt{{b}^{\mathrm{2}} +\mathrm{9}}+\sqrt{{c}^{\mathrm{2}} +\mathrm{16}\:}\right]^{\mathrm{2}} \\ $$$$={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{1}+\mathrm{9}+\mathrm{16}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\sqrt{{b}^{\mathrm{2}} +\mathrm{9}}+ \\ $$$$\mathrm{2}\sqrt{{b}^{\mathrm{2}} +\mathrm{9}}\sqrt{{c}^{\mathrm{2}} +\mathrm{16}}+\mathrm{2}\sqrt{{c}^{\mathrm{2}} +\mathrm{16}}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\geqslant{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{26}+\mathrm{2}\left({ab}+\mathrm{1}×\mathrm{3}\right)+\mathrm{2}\left({bc}+\mathrm{3}×\mathrm{4}\right)+ \\ $$$$\mathrm{2}\left({ac}+\mathrm{4}×\mathrm{1}\right) \\ $$$$=\left({a}+{b}+{c}\right)^{\mathrm{2}} −\mathrm{2}{ab}−\mathrm{2}{bc}−\mathrm{2}{ca}+\mathrm{2}{ab}+\mathrm{2}{ac}+\mathrm{2}{bc}+\mathrm{26} \\ $$$$+\mathrm{6}+\mathrm{24}+\mathrm{8}=\mathrm{15}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} =\mathrm{17}^{\mathrm{2}} \\ $$$${E}^{\mathrm{2}} \geqslant\mathrm{17}^{\mathrm{2}} \Rightarrow{E}\geqslant\mathrm{17} \\ $$$${Equality}\:{when}\:\left({a},{b},{c}\right)=\left(\lambda,\mathrm{3}\lambda,\mathrm{4}\lambda\right)\Rightarrow\mathrm{8}\lambda=\mathrm{15} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{15}}{\mathrm{8}}\Rightarrow\left({a},{b},{c}\right)=\left(\frac{\mathrm{15}}{\mathrm{8}},\frac{\mathrm{45}}{\mathrm{8}},\frac{\mathrm{15}}{\mathrm{2}}\right) \\ $$
Commented by A5T last updated on 24/Jun/24
$$\mid\boldsymbol{{u}}\mid\mid\boldsymbol{{v}}\mid\geqslant\boldsymbol{{u}}\centerdot\boldsymbol{{v}} \\ $$$$\boldsymbol{{u}}=\left({u}_{\mathrm{1}} ,{u}_{\mathrm{2}} \right),\:\boldsymbol{{v}}=\left({v}_{\mathrm{1}} ,{v}_{\mathrm{2}} \right) \\ $$$$\Rightarrow\sqrt{{u}_{\mathrm{1}} ^{\mathrm{2}} +{u}_{\mathrm{2}} ^{\mathrm{2}} }\sqrt{{v}_{\mathrm{1}} ^{\mathrm{2}} +{v}_{\mathrm{2}} ^{\mathrm{2}} }\geqslant{u}_{\mathrm{1}} {v}_{\mathrm{1}} +{u}_{\mathrm{2}} {v}_{\mathrm{2}} \\ $$$${Equality}\:{holds}\:{when}\:\left({u}_{\mathrm{1}} ,{u}_{\mathrm{2}} \right)=\left(\lambda{v}_{\mathrm{1}} ,\lambda{v}_{\mathrm{2}} \right) \\ $$
Commented by Ismoiljon_008 last updated on 24/Jun/24
$${thank} \\ $$$$ \\ $$