Let a triangle with sides a b and c. $p=\frac{a+b+c}{2}$ Prove that $\frac{(p-a)(p-b)(p-c)(a^2+b^2+c^2+ab+bc+ca)}{abcp^2}\le\frac{2}{3}$
My thought: Let $a=x+y,b=y+z,c=z+x$ so $x+y+z=p$ Then let $p=x+y+z,q=xy+yz+zx,r=xyz$ we get $10p^2r\le3qr+p^3q$ I don't know what to do next. Thanks