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If   \(a,b,c,x,y,z\)   are positive and \(x^2 + y^2 + z^2 = 36,\)  and \(ax + by + cz = 30,\) compute

 

\(\frac{a + b + c}{x + y + z}\)
 

 Aug 9, 2019
 #1
avatar+6046 
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\(\text{well.... if this is true for all $a,b,c,x,y,z$, then I can choose $(x,y,z)=2\sqrt{3}(1,1,1)$}\\ (a,b,c)\cdot (x,y,z)=30\\ 2\sqrt{3}(a+b+c)=30\\ (a+b+c) = 5\sqrt{3}\\ x+y+z = 6\sqrt{3}\\ \dfrac{a+b+c}{x+y+z} = \dfrac{5\sqrt{3}}{6\sqrt{3}}=\dfrac 5 6\)

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 Aug 10, 2019

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