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- Feb 14, 2012

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Prove that $\dfrac{1}{3}\le\dfrac{u}{v}\le\dfrac{1}{2}$ and that the fraction $\dfrac{1}{2}$ on the right cannot be replaced by a smaller number.

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- #1

- Feb 14, 2012

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Prove that $\dfrac{1}{3}\le\dfrac{u}{v}\le\dfrac{1}{2}$ and that the fraction $\dfrac{1}{2}$ on the right cannot be replaced by a smaller number.

- Mar 31, 2013

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$=(2(a^2+b^2+c^2 - ab + bc + ca) >=0$

or $a^2 + b^2 + c^2 >= ab + bc + ac \cdots(1)$

Now as a, b, c are sides of triangle we have

$a <= b + c$ or $a^2 <= ab + ac\cdots(2)$

similarly $b^2 <= ab + bc\cdots(3)$

and $c^2 <= bc + ac\cdots(4)$

adding (2) ,(3) , (4) we get $a^2 + b^2 + c^2 <= 2ab + 2bc + 2ca\cdots(2)$

Now we have $v = (a+b+c)^2 = (a^2 + b^2 + c^2 + 2(ab+bc+ca) <= (a^2+b^2+c^2) + 2(a^2+b^2+c^2$

or $ v <= 3(a^2+b^2+ c^2$

or $ v<=3u$

or $\frac{1}{3} < = \frac{u}{v}\cdots(5)$

Now for the 2nd part

$2u= u + u = (a^2+b^2+c^2) + (a^2 + b^2 + c^2 ) <= (a^2+b^2+c^2) + 2(ab+bc+ca)$ using (2)

or $2u <= (a+b+c)^2$ or $2u <= v\cdots(6)$

from (5) and (6) we get the result

Last edited:

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- Feb 14, 2012

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There was a typo in 2nd last line it is corrected

other wise solution is complete

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- Feb 7, 2012

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