MHB Prove $\frac{1}{3}\le \frac{u}{v} \le \frac{1}{2}$ with Triangle Sides

[anemone]

Let $a,\,b$ and $b$ be the lengths of the sides of a triangle. Suppose that $u=a^2+b^2+c^2$ and $v=(a+b+c)^2$.

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.

 

[kaliprasad]

Spoiler: My solution

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

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

 

[anemone]

Hi kaliprasad, thanks for participating! But I was wondering if you have somehow missed out the solution to the very last part of the question...

 

[kaliprasad]

Hi kaliprasad, thanks for participating! But I was wondering if you have somehow missed out the solution to the very last part of the question...

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

Spoiler

I could not complete the part that fraction on the right cannot be replaced by a smaller number

other wise solution is complete

 

[Opalg]

About the last part of the question...

Spoiler

If $a=b$ and $c=0$ then $u=2a^2$ and $v=4a^2$. So $\frac uv = \frac12.$ Of course, you cannot have a triangle with one side zero. But suppose you take an isosceles triangle with sides $a$, $a$ and $\varepsilon$. Then $$2u - v = 4a^2 + 2\varepsilon^2 - (2a + \varepsilon)^2 = \varepsilon(4a - \varepsilon),$$ which you can make as small as you like by taking $\varepsilon$ small enough. So $v$ can be made arbitrarily close to $2u$ and thus $\frac uv$ can be made arbitrarily close to $\frac12.$