den_thewhitelion wrote: Let a,b,c be real numbers such that:$a\ge b\ge 1\ge c\ge 0$ and a+b+c=1. Are you sure?

$a=x+1,b=y+1,c=z;x\geq 0,y\geq 0 ;1\geq z \geq 0;x+y+z=1$ $ab+ac+bc=xy+x+y+1+xz+z+yz+z=xy+xz+yz+z+2=xy+(x+y)(1-x-y)+1-x-y+2=xy-(x+y)^2+3$ $xy-(x+y)^2+3 \leq 3$ - obvious. Equality for $(a,b,c)=(1,1,1)$ $xy+xz+yz+z+2 \geq 2$ - obvious. Equality for $(a,b,c)=(1,2,0),(2,1,0)$

den_thewhitelion wrote: Let $a,b,c$ be real numbers such that:$a\ge b\ge 1\ge c\ge 0$ and $a+b+c=3.$ $a)$ Prove that $2\le ab +bc+ca\le 3$ $b)$ Prove that $\dfrac{24}{a^3+b^3+c^3}+\dfrac{25}{ab+bc+ca}\ge 14$. Find the equality cases. http://www.artofproblemsolving.com/community/c6h1261224p6547636