By Titu's inequality LHS =$\frac{\frac{1}{(a-1)^2}}{a-1}+\frac{\frac{1}{(b-1)^2}}{b-1} +\frac{\frac{1}{(c-1)^2}}{c-1}\ge \frac{(\frac{1}{a-1}+\frac{1}{b-1}+\frac{1}{c-1})^2}{a+b+c-3}$ .We want to prove that $\frac{1}{a-1}+\frac{1}{b-1}+\frac{1}{c-1}> 1$.We assume that is not true.Then $a,b,c>2$.Also,at least two of $a,b,c >3$.==>$ ab+bc+ca>3*3+2*3+2*3=21$. Contradiction.

Misha57 wrote: By Titu's inequality LHS =$\frac{\frac{1}{(a-1)^2}}{a-1}+\frac{\frac{1}{(b-1)^2}}{b-1} +\frac{\frac{1}{(c-1)^2}}{c-1}\ge \frac{\frac{1}{a-1}+\frac{1}{b-1}+\frac{1}{c-1}}{a+b+c-3}$ .We want to prove that $\frac{1}{a-1}+\frac{1}{b-1}+\frac{1}{c-1}> 1$.We assume that is not true.Then $a,b,c>2$.Also,at least two of $a,b,c>3$.$\rightarrow$ $ab+bc+ca>3*3+2*3+2*3=21$. Contradiction. Note that $\frac{\frac{1}{a-1} +\frac{1}{b-1} +\frac{1}{c-1}}{a+b+c-3}$ should be $\frac{(\frac{1}{a-1} +\frac{1}{b-1} +\frac{1}{c-1})^2}{a+b+c-3}$ And Titu inequality only work in the case $t-1 \geq 0$ for all $t=a,b,c$ Why $\frac{1}{a-1} +\frac{1}{b-1} +\frac{1}{c-1} \leq 1$ implie $a,b,c >2$ and two of $a,b,c$ $>3$ What about $a=b=c=0.5$

ThE-dArK-lOrD wrote: Misha57 wrote: By Titu's inequality LHS =$\frac{\frac{1}{(a-1)^2}}{a-1}+\frac{\frac{1}{(b-1)^2}}{b-1} +\frac{\frac{1}{(c-1)^2}}{c-1}\ge \frac{\frac{1}{a-1}+\frac{1}{b-1}+\frac{1}{c-1}}{a+b+c-3}$ .We want to prove that $\frac{1}{a-1}+\frac{1}{b-1}+\frac{1}{c-1}> 1$.We assume that is not true.Then $a,b,c>2$.Also,at least two of $a,b,c>3$.$\rightarrow$ $ab+bc+ca>3*3+2*3+2*3=21$. Contradiction. Note that $\frac{\frac{1}{a-1} +\frac{1}{b-1} +\frac{1}{c-1}}{a+b+c-3}$ should be $\frac{(\frac{1}{a-1} +\frac{1}{b-1} +\frac{1}{c-1})^2}{a+b+c-3}$ And Titu inequality only work in the case $t-1 \geq 0$ for all $t=a,b,c$ Why $\frac{1}{a-1} +\frac{1}{b-1} +\frac{1}{c-1} \leq 1$ implie $a,b,c >2$ and two of $a,b,c$ $>3$ What about $a=b=c=0.5$ $a,b,c>1$

zschess wrote: Let $ a, b, c $ be the side length of a triangle, with $ ab + bc + ca = 18 $ and $ a, b, c > 1 $. Prove that $$ \sum_{cyc}\frac{1}{(a - 1)^3} > \frac{1}{a + b + c - 3} $$ Let $a=1+x,b=1+y,c=1+z,x,y,z>0$,then,the inequality becomes the following problem： Let $x,y,z>0$ be positive real numbers such that $2(x+y+z)+xy+yz+zx=15$,Prove that \[\sum{\frac{1}{x^3}}\ge{\frac{1}{x+y+z}}\] Indeed,we have \[\sum{\frac{1}{x^3}}-\frac{1}{x+y+z}=\frac{1}{18x^2y^2z^2(x+y+z)}\sum{\frac{(3x^2y^2z^2+x^2yz^3+xy^2z^3+6x^2y^2+6xz^3+6yz^3)(x-y)^2}{xy}}+\]\[\frac{2(x^4y^3z+x^4yz^3+x^3y^4z+x^3y^3z^2+x^3y^2z^3+x^3yz^4+x^2y^3z^3+xy^4z^3+xy^3z^4+3x^4y^3+3x^4z^3+3x^3y^4+3x^3z^4+3y^4z^3+3y^3z^4)}{9x^3y^3z^3(x+y+z)}\]\[\ge{0}\]

szl6208 wrote: zschess wrote: Let $ a, b, c $ be the side length of a triangle, with $ ab + bc + ca = 18 $ and $ a, b, c > 1 $. Prove that $$ \sum_{cyc}\frac{1}{(a - 1)^3} > \frac{1}{a + b + c - 3} $$ Let $a=1+x,b=1+y,c=1+z,x,y,z>0$,then,the inequality becomes the following problem： Let $x,y,z>0$ be positive real numbers such that $x+y+z+2(xy+yz+zx)=15$,Prove that \[\sum{\frac{1}{x^3}}\ge{\frac{1}{x+y+z}}\] Indeed,we have \[\sum{\frac{1}{x^3}}-\frac{1}{x+y+z}=\frac{1}{18x^2y^2z^2(x+y+z)}\sum{\frac{(3x^2y^2z^2+x^2yz^3+xy^2z^3+6x^2y^2+6xz^3+6yz^3)(x-y)^2}{xy}}+\]\[\frac{2(x^4y^3z+x^4yz^3+x^3y^4z+x^3y^3z^2+x^3y^2z^3+x^3yz^4+x^2y^3z^3+xy^4z^3+xy^3z^4+3x^4y^3+3x^4z^3+3x^3y^4+3x^3z^4+3y^4z^3+3y^3z^4)}{9x^3y^3z^3(x+y+z)}\]\[\ge{0}\] But $ab+bc+ca=18=(x+1)(y+1)+(y+1)(z+1)+(z+1)(x+1)=x+y+xy+1+y+z+yz+1+z+x+zx+1=2(x+y+z)+xy+yz+zx+3$, so $2(x+y+z)+xy+yz+zx=15$ but not $2(xy+yz+zx)+x+y+z=15$

Why did the condition that a,b,c were the sides of a triangle ever matter?

zschess wrote: Let $ a, b, c $ be the side length of a triangle, with $ ab + bc + ca = 18 $ and $ a, b, c > 1 $. Prove that $$ \sum_{cyc}\frac{1}{(a - 1)^3} > \frac{1}{a + b + c - 3} $$ WLOG $ca=min\{ab,bc,ca\}.$ If $ca<2,$ hence $1<c,a<2,b<c+a<4,a+b+c<12;$ If $ca<2,$ hence $ab,bc>2,b^2<b(c+a)=18ca<16,b<4,$ similarly $c<4,a<4,$ so $a+b+c<12.$ $0<a+b+c-3<9,$ $\sum_{cyc}\frac{1}{(a - 1)^3} \geq\frac{3}{(a - 1)(b - 1)(c - 1)} \geq\frac{81}{(a+b+c - 3)^3}>\frac{1}{a+b+c - 3}.$

The following inequality is also true

luofangxiang wrote: The following inequality is also true Let $a,b,c>1$, $\sum_{cyc}ab=18$; Prove: $\sum_{cyc}\frac{1}{(a-1)^3}\geq\left(\frac{3}{\sqrt6-1}\right)^2\frac{1}{\sum\limits_{cyc}a-3}$ Let $a=x+1$, $b=y+1$ and $c=z+1$. Hence, $x$, $y$ and $z$ are positives such that $xy+xz+yz+2(x+y+z)=15$ and we need to prove that $\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\geq\frac{\left(\frac{3}{\sqrt6-1}\right)^2}{x+y+z}$. Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$. Thus, the condition does not depend on $w^3$ and since $x^3y^3+x^3z^3+y^3z^3=27v^6-27uv^2w^3+3w^6$ is a decreasing function of $w^3$, it remains to prove our inequality for a maximal value of $w^3$, which happens for equality case of two variables. Let $y=x$. Hence, $z=\frac{15-4x-x^2}{2(x+1)}$, which gives that your inequality is true.

arqady Thank you very much