Given triangle sides $a$, $b$, $c$, show $\sum_{\text{cyc}}\frac{a}{a+b}\geq\frac32\prod_{\text{cyc}}(\frac{a}{a+b}+\frac{b}{b+c})$

Question

If $a,b,c$ are three sides of a triangle then I need to prove that $$ {\frac {a}{a+b}}+{\frac {b}{b+c}}+{\frac {c}{c+a}}\geq \frac{3}{2}\, \left( { \frac {a}{a+b}}+{\frac {b}{b+c}} \right) \left( {\frac {b}{b+c}}+{ \frac {c}{c+a}} \right) \left( {\frac {c}{c+a}}+{\frac {a}{a+b}} \right) $$

My Try:

I know that in a triangle $a + b > c$ so I have $\frac{1}{a+b} < \frac{1}{c}$ which implies that $\frac{a}{a+b} < \frac{a}{c}$ and so on, using which I conclude that

$RHS < \frac{3}{2}(\frac{a}{c} + \frac{b}{a})(\frac{b}{a} + \frac{c}{b})(\frac{c}{b} + \frac{a}{c})$

Now I need to prove that the RHS of the above inequality is less than the LHS of the inequality mentioned in the question $${\frac {a}{a+b}}+{\frac {b}{b+c}}+{\frac {c}{c+a}} > \frac{3}{2}(\frac{a}{c} + \frac{b}{a})(\frac{b}{a} + \frac{c}{b})(\frac{c}{b} + \frac{a}{c})$$

I now substitute $\frac{b}{a} = x, \frac{a}{c} = y, \frac{c}{b} = z$ and the above inequality, if I am correct simplifies to

$$ \frac{x}{x+1} + \frac{y}{y+1} + \frac{z}{z+1} > \frac{3}{2}(x+y)(y+z)(z+x)$$

Now how do I prove the above inequality, moreover how do I prove the equality case?

Answer 1

Let $a=y+z$, $b=x+z$ and $c=y+z$.

Thus, $x=\frac{b+c-a}{2}>0$ and similarly $y>0$ and $z>0$.

Thus, after full expending we need to prove that $$\sum_{cyc}(3x^6+13x^5y+5x^5z+16x^4y^2-10x^4z^2-3x^3y^3+15x^4yz-7x^3y^2z-23x^3z^2y-9x^2y^2z^2)\geq0$$ or $$\sum_{cyc}(4x^5y+x^4y^2-3x^3y^3-10x^2y^4+5xy^4+3y^6)+$$ $$+\sum_{cyc}(9x^5y+15x^4y^2+15x^4yz-7x^3y^2z-23x^3z^2y-9x^2y^2z^2)\geq0$$ or $$\sum_{cyc}y(x-y)^2(4x^3+9x^2y+11xy^2+3y^3)+9\sum_{cyc}(x^5y-x^2y^2z^2)+$$ $$+7xyz\sum_{cyc}(x^3-x^2y)+8xyz\sum_{cyc}(x^3-x^2z)+15\sum_{cyc}(x^4y^2-x^3z^2y)\geq0,$$ which is true by AM-GM and Rearrangement.

Also, we can use the $uvw$.

Indeed, let $\frac{b}{a}=x$, $\frac{c}{b}=y$, $\frac{a}{c}=z$.

Thus, $xyz=1$ and we need to prove that $$\sum_{cyc}\frac{1}{1+x}\geq\frac{3}{2}\prod_{cyc}\left(\frac{1}{1+x}+\frac{1}{1+y}\right)$$ or $$2\prod_{cyc}(1+x)\sum_{cyc}(xy+2x+1)\geq3\prod_{cyc}(x+y+2)$$ or $$\sum_{cyc}(2x^2y^2+3x^2y+3x^2z-2x^2-6x)\geq0$$ or $$2(9v^4-6uw^3)+3(9uv^2-3w^3)w-18uw^3\geq0$$ or $f(v^2)\geq0,$ where $$f(v^2)=6v^4+9uv^2w-6u^2w^2+4v^2w^2-10uw^3-3w^4.$$ But it's obvious that $f$ increases, which says that it's enough to prove our inequality for the minimal value of $v^2$, which happens in the following cases.

1. Sum of two sides-lengths of the triangle closes to the third.

Let $x\geq y\geq z$.

Thus, $x\geq1$, $b=ax$, $c=by$ and $c=xya$.

Now, $b+c>a$ gives $x+xy>1,$ which is obvious.

$a+c>b$ gives $1+xy>x$ and we need to check $y\rightarrow\frac{x-1}{x}$, $z\rightarrow\frac{1}{x-1},$ where $x>1$, or for $x=1+t$ we have $y\rightarrow\frac{t}{t+1}$ and $z\rightarrow\frac{1}{t},$ which gives $$3t^6+5t^5-10t^4-3t^3+16t^2+13t+3>0,$$ which is obvious.

Also, we have $a+b>c$, which gives $1+x>xy$ and we need to check $y\rightarrow \frac{1+x}{x}$ and $z\rightarrow\frac{1}{1+x},$ which gives $$3x^6+13x^5+16x^4-3x^3-10x^2+5x+3\geq0,$$ which was already for $x=\frac{1}{t};$

2. Two variables are equal.

Let $y=x$.

Thus, $z=\frac{1}{x^2}$ and $a+b>c$ gives $$a+ax>x^2a$$ or $$x<\frac{1+\sqrt5}{2}.$$ Also, $b+c>a$ gives $$ax+x^2a>a$$ or $$x>\frac{\sqrt5-1}{2}$$ and we obtain $$\frac{\sqrt5-1}{2}<x<\frac{1+\sqrt5}{2}.$$ Id est, it remains to prove that $$\frac{2}{1+x}+\frac{1}{1+\frac{1}{x^2}}\geq\frac{3}{2}\left(\frac{1}{1+x}+\frac{1}{1+\frac{1}{x^2}}\right)^2\frac{2}{1+x}$$ or $$(x-1)^2(x^5+4x^4+3x^3+2x-1)\geq0,$$ which is obviously true for $\frac{\sqrt5-1}{2}<x<\frac{1+\sqrt5}{2}.$

Done again!