parmenides51 wrote: Prove that if positive integers $x,y$ satisfy $x+y= 3$, $x,y \ge 1$ then $$9(x- 1)(y- 1) + (y^2 + y+ 1)(x + 1) + (x^2-x- 1)(y- 1) \ge 9$$ $$?$$

Let $x,y\geq 1 $ and $x+y= 3.$ Prove that$$ (y^2 + y+ 1)(x + 1) + (x^2-x- 1)(y- 1) \ge 9$$

sqing wrote: parmenides51 wrote: Prove that if positive integers $x,y$ satisfy $x+y= 3$, $x,y \ge 1$ then $$9(x- 1)(y- 1) + (y^2 + y+ 1)(x + 1) + (x^2-x- 1)(y- 1) \ge 9$$ $$?$$ Sorry about the typo it should be $9(x- 1)(y- 1) + (y^2 + y+ 1)(x + 1) + (x^2-x+ 1)(y- 1) \ge 9$

sqing wrote: Let $x,y\geq 1 $ and $x+y= 3.$ Prove that$$ (y^2 + y+ 1)(x + 1) + (x^2-x- 1)(y- 1) \ge 9$$ Let $x-y=k$ then $x=\frac{k+3}{2}$ and $y=\frac{3-k}{2}$ which means that $-1 \le k \le 1$ So now the ineq after simplifying becomes: $$(k^2-8k+19)(k+5)+(k^2+4k-1)(1-k) \ge 72$$And after expanding and canceling terms the ineq is $$-6k^2-16k+94 \ge 72 \implies 6k^2+16k \le 22$$But the last ineq is true since $k \le 1$ hence the first ineq is also true and we are done Smol detail: so equality holds when $k=1$ meaning that on the first one the equality holds when $y=1,x=2$

Let $x,y\geq 1 $ and $x+y= 3.$ Prove that$$ (y^2 + y+ 1)(x + 1) + (x^2-x- 2)(y- 2) \ge 9$$

sqing wrote: Let $x,y\geq 1 $ and $x+y= 3.$ Prove that$$ (y^2 + y+ 1)(x + 1) + (x^2-x- 2)(y- 2) \ge 9$$ $$ (y^2 + y+ 1)(x + 1) + (x^2-x- 2)(y- 2)=-4x^2+7x+11$$$$=9-(x-2)(4x+1)= 9+(y-1)(4x+1) \ge 9$$

parmenides51 wrote: Prove that if positive integers $x,y$ satisfy $x+y= 3$, $x,y \ge 1$ then $$9(x- 1)(y- 1) + (y^2 + y+ 1)(x + 1) + (x^2-x+ 1)(y- 1) \ge 9$$ Only possibilities for positive integers $\ge 1$ whose sum is $3$ are $(x,y)=(1,2)$ and $(x,y)=(2,1)$ So just two immediate checks to do in order to get the required result.

parmenides51 wrote: Prove that if positive integers $x,y$ satisfy $x+y= 3$, $x,y \ge 1$ then $$9(x- 1)(y- 1) + (y^2 + y+ 1)(x + 1) + (x^2-x+ 1)(y- 1) \ge 9$$ Note $(x,y)=(1,2)$ or $(x,y)=(2,1)$. If $(x,y)=(1,2)$, then the $LHS$ is $(2^2+2+1)(1+1)+1=15\ge 9$. If $(x,y)=(2,1)$, then the $LHS$ is $(1^2+1+1)(2+1)=3\cdot 3=9\ge 9$.

pco wrote: parmenides51 wrote: Prove that if positive integers $x,y$ satisfy $x+y= 3$, $x,y \ge 1$ then $$9(x- 1)(y- 1) + (y^2 + y+ 1)(x + 1) + (x^2-x+ 1)(y- 1) \ge 9$$ Only possibilities for positive integers $\ge 1$ whose sum is $3$ are $(x,y)=(1,2)$ and $(x,y)=(2,1)$ So just two immediate checks to do in order to get the required result. Whoops, typo again. edit: fixed

parmenides51 wrote: Prove that if positive reals $x,y$ satisfy $x+y= 3$, $x,y \ge 1$ then $$9(x- 1)(y- 1) + (y^2 + y+ 1)(x + 1) + (x^2-x+ 1)(y- 1) \ge 9$$ $$9(x- 1)(y- 1) + (y^2 + y+ 1)(x + 1) + (x^2-x+ 1)(y- 1) \ge$$$$ (y^2 + y+ 1)(x + 1) + (x^2-x+ 1)(y- 1) =9+3(x+1)(y-1) \ge 9$$

what a silly one Anyway just test $(1,2),(2,1)$ and both of these satisfy this inequality $\blacksquare$