If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is:

Question

If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is:

$A)$ $\le3$ , $B)$ $\ge3$, $(C)$ $\ge2$, $(D)$ $\le2$

My attempt is as follows:-

$$\dfrac{1}{2}\left(\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}\right)$$

Let $y=\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}$

$$A.M\ge H.M$$ $$\dfrac{\dfrac{s-a}{a}+\dfrac{s-b}{b}+\dfrac{s-c}{c}}{3}\ge \dfrac{3}{y}$$ $$\dfrac{\dfrac{s\cdot (ab+bc+ca)}{abc}-3}{3}\ge\dfrac{3}{y}$$ $$y\ge\dfrac{9}{\dfrac{(a+b+c)(ab+bc+ca)}{2abc}-3}\tag{1}$$

Let $z=\dfrac{(a+b+c)(ab+bc+ca)}{abc}$

Equation $(1)$ will give us $y_{min}$, so for that we need to find maximum value of $z$

But unfortunately I was able to find minimum value of $z$ in the following way

$$\dfrac{a+b+c}{3}\ge \dfrac{3abc}{ab+bc+ca}$$ $$\dfrac{(a+b+c)(ab+bc+ca)}{abc}\ge 9$$

But nevertheless, I tried plugging this minimum value of $z$ into equation $(1)$ and I got $y\ge 6$ and as the original expression was $\dfrac{y}{2}$, so $\dfrac{y}{2}\ge 3$ and surprisingly this answer is correct. What am I missing here?

Answer 1

If $x,y,z>0$, then the sides of any triangle are : Then $$a=y+z, b=x+z,c=x+y \implies b+c-a=2x, c+a-b=2y, a+b-c=2x$$ Then the LHS of the required inequality becomes: $$\frac{y+z}{2x}+\frac{x+z}{2y}+\frac{x+y}{2z}=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}\right)\ge \frac{1}{2}(2+2+2)\ge 3$$ Here we have used $AM+GM: P+Q\ge 2\sqrt{PQ}$. The equality holds when the triangle is equilateral.

Answer 2

Note that $f(x)=\frac x{s-x}$ is a convex function. Then, Jensen’s inequality leads to,

$$\begin{align} & \dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}\\ =& \frac12 \left(\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}\right) \\ \ge &\frac12\cdot3\cdot\frac{ \frac{a+b+c}3}{s -\frac{a+b+c}3}=\frac32\cdot\frac{\frac{2s}3}{s-\frac{2s}3}=3 \end{align} $$

Answer 3

By C-S $$\sum_{cyc}\frac{a}{b+c-a}=\sum_{cyc}\frac{a^2}{a(b+c-a)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}a(b+c-a)}\geq3,$$ where the last inequality it's just $$\sum_{cyc}(a-b)^2\geq0.$$

Answer 4

The answer is B). See my solution.

By your work we need to prove that: $$\dfrac{9}{\dfrac{(a+b+c)(ab+bc+ca)}{2abc}-3}\tag{1}\geq6,$$ which is impossible because it's wrong (you got a reversed inequality).

Another way.

$$\sum_{cyc}\frac{a}{b+c-a}-3=\sum_{cyc}\left(\frac{a}{b+c-a}-1\right)=\sum_{cyc}\frac{a-b-(c-a)}{b+c-a}=$$ $$=\sum_{cyc}(a-b)\left(\frac{1}{b+c-a}-\frac{1}{c+a-b}\right)=\sum_{cyc}\frac{2(a-b)^2}{(b+c-a)(c+a-b)}\geq0.$$

Answer 5

I don't see what's wrong with your answer, but it is a long method that gives an incorrect answer, so I'd suggest another way, such as the following. By scaling we can assume that $a + b + c = 1$, in which case we want to find a lower bound for $${a \over 1- 2a} + {b \over 1 - 2b} + {c \over 1 - 2c} $$ $$-{3 \over 2} + {1 \over 2}\bigg({1 \over 1- 2a} + {1 \over 1 - 2b} + {1 \over 1 - 2c}\bigg)$$ Apply the AM-HM inequality in the expression in parentheses and the above is greater than or equal to $$-{3 \over 2} + {9 \over 2}\bigg({ 1- 2a} + { 1 - 2b} + { 1 - 2c}\bigg)^{-1}$$ Since $a + b + c = 1$ this is exactly equal to $3$, which is achieved when $a = b = c = {1 \over 3}$.