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Question Number 204621 by hardmath last updated on 23/Feb/24
a , b , c ∈ R^+ If (√a) + (√b) + (√c) = 1 Prove that: a + b + c ≥ (1/3)
$$\mathrm{a}\:,\:\mathrm{b}\:,\:\mathrm{c}\:\in\:\mathbb{R}^{+} \\ $$$$\mathrm{If}\:\:\:\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{b}}\:+\:\sqrt{\mathrm{c}}\:=\:\mathrm{1} \\ $$$$\mathrm{Prove}\:\mathrm{that}:\:\:\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Answered by A5T last updated on 23/Feb/24
((a+b+c)/3)≥((((√a)+(√b)+(√c))/3))^2 =(1/9)⇒a+b+c≥(1/3)
$$\frac{{a}+{b}+{c}}{\mathrm{3}}\geqslant\left(\frac{\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{9}}\Rightarrow{a}+{b}+{c}\geqslant\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Answered by mr W last updated on 23/Feb/24
a+b+c=((((√a))^2 )/1)+((((√b))^2 )/1)+((((√c))^2 )/1) ≥((((√a)+(√b)+(√c))^2 )/(1+1+1))=(1^2 /3)=(1/3)
$${a}+{b}+{c}=\frac{\left(\sqrt{{a}}\right)^{\mathrm{2}} }{\mathrm{1}}+\frac{\left(\sqrt{{b}}\right)^{\mathrm{2}} }{\mathrm{1}}+\frac{\left(\sqrt{{c}}\right)^{\mathrm{2}} }{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\geqslant\frac{\left(\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\right)^{\mathrm{2}} }{\mathrm{1}+\mathrm{1}+\mathrm{1}}=\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by mr W last updated on 23/Feb/24
Answered by mr W last updated on 24/Feb/24
Commented by mr W last updated on 24/Feb/24
say (√a)=x, (√b)=y, (√c)=z x+y+z=1 is the plane ΔABC. P(x,y,z) is a point on the plane. a+b+c=x^2 +y^2 +z^2 =OP^2 OP_(min) is the distance d from O to the plane ΔABC. OP_(min) =d=((∣0+0+0−1∣)/( (√(1^2 +1^2 +1^2 ))))=(1/( (√3))) OP_(min) ^2 =(1/3) ⇒a+b+c=OP^2 ≥OP_(min) ^2 =(1/3)
$${say}\:\sqrt{{a}}={x},\:\sqrt{{b}}={y},\:\sqrt{{c}}={z} \\ $$$${x}+{y}+{z}=\mathrm{1}\:{is}\:{the}\:{plane}\:\Delta{ABC}. \\ $$$${P}\left({x},{y},{z}\right)\:{is}\:{a}\:{point}\:{on}\:{the}\:{plane}. \\ $$$${a}+{b}+{c}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={OP}^{\mathrm{2}} \\ $$$${OP}_{{min}} \:{is}\:{the}\:{distance}\:{d}\:{from}\:{O}\:{to}\:{the} \\ $$$${plane}\:\Delta{ABC}. \\ $$$${OP}_{{min}} ={d}=\frac{\mid\mathrm{0}+\mathrm{0}+\mathrm{0}−\mathrm{1}\mid}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${OP}_{{min}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{a}+{b}+{c}={OP}^{\mathrm{2}} \geqslant{OP}_{{min}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by hardmath last updated on 24/Feb/24
Perfect solution, thank you dear professor
$$\mathrm{Perfect}\:\mathrm{solution},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$

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