Triangle Challenge: Proving Inequality of Sides

[anemone]

Let $a\,b$ and $c$ be the sides of a triangle.

Prove that \(\displaystyle \frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\ge 3\).

 

[Greg]

Let $a\,b$ and $c$ be the sides of a triangle.

Prove that \(\displaystyle \frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\ge 3\).

Spoiler

Let $A=\dfrac{a}{b+c-a},B=\dfrac{b}{a+c-b},C=\dfrac{c}{a+b-c}$.

By the relationship between the arithmetic mean and the harmonic mean we have
$$\dfrac{A+B+C}{3}\ge\dfrac{3}{\dfrac1A+\dfrac1B+\dfrac1C}\Rightarrow A+B+C\ge\dfrac{9}{\dfrac1A+\dfrac1B+\dfrac1C}$$

The rational expression on the RHS is at a maximum when the denominator is at a minimum.
$$\dfrac{b+c-a}{a}+\dfrac{a+c-b}{b}+\dfrac{a+b-c}{c}$$
$$=\dfrac{b+c}{a}-1+\dfrac{a+c}{b}-1+\dfrac{a+b}{c}-1$$
$$=\dfrac ba+\dfrac ab+\dfrac ca+\dfrac ac+\dfrac cb+\dfrac bc-3\quad(1)$$
$$(x-y)^2\ge0$$
$$x^2+y^2\ge2xy$$
$$\dfrac xy+\dfrac yx\ge2$$
hence the minimum of $(1)$ is $3$, so we have
$$A+B+C\ge\dfrac{9}{\dfrac1A+\dfrac1B+\dfrac1C}\implies A+B+C\ge3$$

$\text{Q.E.D.}$

 

[kaliprasad]

Let $a\,b$ and $c$ be the sides of a triangle.

Prove that \(\displaystyle \frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\ge 3\).

Spoiler

if we put $x = b+c-a, y = a+c-b, z = a + b -c$ we get
Given expression
$= \dfrac{1}{2}(\dfrac{y+z}{x} +\dfrac{z+x}{y} + \dfrac{x+y}{z})$
$= \dfrac{1}{2}((\dfrac{y}{x}+ \dfrac{x}{y})+( \dfrac{z}{y} + \dfrac{y}{z}) + ( \dfrac{z}{x} + \dfrac{x}{z}))$
$= \dfrac{1}{2}(((\sqrt{\frac{y}{x}}- \sqrt{\frac{x}{y}})^2+ 2) +((\sqrt{\frac{y}{z}}- \sqrt{\frac{z}{y}})^2+ 2) + ((\sqrt{\frac{z}{x}}- \sqrt{\frac{x}{z}})^2+ 2)))$
$\ge \dfrac{1}{2}(2+2+2)\,or\,3$

 

[anemone]

Thanks both for participating! I actually solved it differently and I will post my solution when I have the time...

By the way, I see that direct use of the AM-GM inequality could bring us to the proof right away in both of the solutions above...:D

Spoiler

Since $a,\,b,\,c>0$ then we have

$=\dfrac ba+\dfrac ab+\dfrac ca+\dfrac ac+\dfrac cb+\dfrac bc-3$

\(\displaystyle \ge 6\sqrt[6]{\dfrac ba\cdot \dfrac ab \cdot \dfrac ca \cdot \dfrac ac \cdot \dfrac cb \cdot \dfrac bc}-3\)

$\ge 6-3=3$

Spoiler

From the substitutions that define $x = b+c-a, y = a+c-b, z = a + b -c$ and since $a,\,b$ and $c$ are sides of triangle, we have $x,\,y,\,z\ge 0$ so that

$ \dfrac{1}{2}((\dfrac{y}{x}+ \dfrac{x}{y})+( \dfrac{z}{y} + \dfrac{y}{z}) + ( \dfrac{z}{x} + \dfrac{x}{z}))$

$\ge \dfrac{1}{2}\left(6\sqrt[6]{\dfrac{y}{x}\cdot \dfrac{x}{y} \cdot \dfrac{z}{y} \cdot \dfrac{y}{z} \cdot \dfrac{z}{x} \cdot \dfrac{x}{z}}\right)$

$\ge 3$

 

[anemone]

My solution:

Spoiler

\(\displaystyle \begin{align*}\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}&=\frac{a^2}{a(b+c)-a^2}+\frac{b^2}{b(a+c)-b^2}+\frac{c^2}{c(a+b)-c^2}\\&\ge 4\left(\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{c+a}\right)^2+\left(\frac{c}{a+b}\right)^2\right)\,\,\text{since}\,\,a^2+\frac{(b+c)^2}{4}\ge a(b+c)\\&\ge 4\left(\frac{\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2}{3}\right)\,\,\text{since}\,\,3(x^2+y^2+z^2)\ge (x+y+z)^2\\&\ge 4\left(\frac{\left(\frac{3}{2}\right)^2}{3}\right)\,\,\text{from the Nesbitt's inequality}\\&\ge 3\end{align*}\)