Prove that: $\sum\frac{x^3}{y+2}+2\ge \sum{x^2}$

Question

Let $x, y, z\ge 0$ such that: $x+y+z=3$. Prove that: $\frac{x^3}{y+2}+\frac{y^3}{z+2}+\frac{z^3}{x+2}+2\ge x^2+y^2+z^2$

It's a hard equality .... :( And I need help now :(

Answer 1

this is a brute force method which looks ugly but useful:

WLOG,let $x=Min[x,y,z],y=x+3u,z=x+3v,u,v \ge 0,0 \le u+v \le 1, a=u+v,1 \ge a \ge u$, then $x=1-a,y=1-a+3u,z=1+2a-3u$, put this in and we get :

$243u^5+243(2-4a)u^4+324(4a^2-3a+1)u^3+9(11-19a+90a^2-80a^3)u^2+9a(27a^3-36a^2-17a-11)u+(-27a^5+54a^4+70a^3+33a^2) =243u^3(u^2+2(1-2a)u+(1-2a)^2)+81(4a^2+1)u^3+9(11-19a+90a^2-80a^3)u^2+9a(27a^3-36a^2-17a-11)u+(-27a^5+54a^4+70a^3+33a^2) \ge 0$

$243u^3(u^2+2(1-2a)u+(1-2a)^2) \ge 0$ is trivial and when $u=0$ get "=",so remain

$f(u)=tu^3+pu^2-qu+c \ge 0$, where : $1\ge u \ge 0$

$t=81(4a^2+1) >0\\ p=9(11-19a+90a^2-80a^3) >0\\ q=-9a(27a^3-36a^2-17a-11) \ge 0\\ c=(-27a^5+54a^4+70a^3+33a^2) \ge 0$

$f(u)$ have one local max and one local min point which can be got from :

$f'(u)=0 \implies 3tu^2+2pu-q=0,u_1=\dfrac{-p-\sqrt{p^2+3qt}}{3t}<0,u_2=\dfrac{-p+\sqrt{p^2+3qt}}{3t}>0$

$f_m=c+u_m(tu_m^2+pu_m-q)=c-u_m^2(2tu_m+p)$

so it is clear that $u_1$ is local max point and $u_2$ is local min point.

edit:(correct a big mistake ) now we will prove $f_{min}=c-u_2^2(2tu_2+p) \ge 0 \iff c-\dfrac{\sqrt{p^2-3qt}*(6qt-2p^2)-9pqt+2p^3}{27t^2} \ge 0 \iff 27 c^2 t^2+4 c p^3-18 c p q t-p^2 q^2+4 q^3 t \iff a^2(3857760a^{12}-35650800a^{11}+85588120a^{10}-65930520a^9-38371332a^8+62252304a^7+19681779a^6-11645742a^5+13855565a^4-6491628*a^3+1441473*a^2-274186*a+43923) \ge 0 \iff 87.83a^{12}-811.67a^{11}+1948.6a^{10}-1501a^9-873.6a^8+1417.3a^7+448.1a^6-265.14a^5+315.45a^4-147.8a^3+32.82a^2-6.24a+1 \ge 0 \iff 87.8a^{10}(a-1)^2+636.1a^{10}(1-a)+(1224.1a^{10}-1501a^9+812.8a^8)+(-812.8a^8-873.6a^8+1417.3a^7+215.1a^6)+(233a^6-265.2a^5+75.8a^4)+(239.6a^4-147.8a^3+22.8a^2)+(10a^2-6.3a+1) >0$

the last one is true for $1\ge a \ge 0$ and only when $a=0$ get "$=$"

so when $a=u=0 \implies x=y=z=1$,the "=" is hold.

the last step is to prove : $ u_2<1 \iff \dfrac{-p+\sqrt{p^2+3qt}}{3t}<1 \iff 3t+2p-q>0 \iff 9(27 a^4-196 a^3+271 a^2-49 a+49)>0$

which is trivial also.

QED.