Question

If a, b, c are the sides of a triangle, then $\frac{a}{c+a-b}+\frac{b}{a+b-c}+\frac{c}{b+c-a}\ge 3$

• A. True
• B. False

Answer 1

The correct option is A True

Since a, b, c are the sides of the triangle we have a, b, c > 0 and we know that the sum of any two sides of a triangle is greater than the third side.
$\Rightarrow$ a+b-c, b+c-a, c+a-b are all positive.
Applying $AM\ge GM$ to the quantities.
a b c
c+a -b , a+b-c, b+c-a
we have $\frac{a}{c+a-b}+\frac{b}{a+b-c}+\frac{c}{b+c-a}$
$⩾3\sqrt[3]{3\frac{abc}{(c+a-b)(a+b-c)(b+c-a)}}$
We have
$a^2⩾a^2-(b-c{)}^2$
or $a^2⩾(a+b-c)(a-b+c)$
Similarly
$b^2⩾(b+c-a)(b-c+a)$
$c^2⩾(c+a-b)\left)\right(c-a+b\left)\right)$
Multiplying together we get
$a^2{b}^2{c}^2⩾(a+b-c{)}^2(b+c-a{)}^2(c+a-b{)}^2$
Taking he positive root both sides we have
$abc⩾(a+b-c)(b+c-a)(c+a-b)$
$\Rightarrow \frac{abc}{(a+b-c)(b+c-a)(c+a-b)}⩾1$
$\Rightarrow \sqrt[3]{3\frac{abc}{(a+b-c)(b+c-a)(c+a-b)}}⩾1$
$\Rightarrow \frac{a}{c+a-b}+\frac{b}{a+b-c}+\frac{c}{b+c-a}⩾3$