inequality :$x^2+y^2+z^2=1$ , $(1-xy)(1-yz)(1-zx)\ge\frac{8} {27}$ and $a+b+c=1$,$a^2+b^2+c^2 +3abc \ge \frac {4}{9}$ 1.Let $x$,$y$,$z$ be nonnegetive reals such that $x^2+y^2+z^2=1$.
$(1-xy)(1-yz)(1-zx)\ge\frac{8} {27}$ 


*let $a$,$b$,$c$ be positive reals with $a+b+c=1$ . Prove that $a^2+b^2+c^2 +3abc \ge \frac {4}{9}$.

 A: *

*For the first one: 


Let $p=x+y+z, q= xy+yz+zx, r=xyz$ Which are greater or equal to zero. 
$x^2+y^2+z^2=p^2-2q \rightarrow p^2-2q=1 $ (1)
$\rightarrow (1-xy)(1-yz)(1-zx)=1-q+pr-r^2\geq \frac{8}{27}$ (2)
$p^3 - 4pq +9r \geq 0 \rightarrow p(p^2 -4q)+9r\geq 0 \rightarrow 9r \geq p(2q-1)$  (3)
Also $p^2 \geq 3q $ and $p^2-2q=1$ thus $2q +1 \geq 3q \rightarrow q \leq 1$ (4)
According to above and $ pq -9r \geq 0$ we have 
$p\geq pq \geq 9r \rightarrow p-r \geq \frac{8}{9}p$ 
Thus we have $r(p − r) \geq \frac{8}{9}pr \geq \frac{8}{9}p \frac{p(2q-1)}{9}=\frac{8(2q+1)(2q-1)}{81}$ (5)
with (2) and (5) we have :
$1-q +\frac{8(2q+1)(2q-1)}{81}\geq \frac{8}{27} \rightarrow (1-q)(49-32q)\geq 0$ which is true due to (4)



*Normalize as follow: $9(a+b+c)(a^2 +b^2 +c^2)+27abc \geq 4(a+b+c)^3$
$\rightarrow  5(a^3 +b^3 +c^3) + 3abc \geq 3(ab(a+b) + bc(b+c)+ca(c+a))$
If you know Schur's inequality: 
$a^3 +b^3 +c^3 + 3abc \geq (ab(a+b) + bc(b+c)+ca(c+a))  (1) $ 
and by Muirhead theorem we have :
$4(a^3 +b^3 +c^3) \geq 2(ab(a+b) + bc(b+c)+ca(c+a)) (2)$
by adding (1) and (2) we have the result. 
