If $x,y,z>0: x+y+z=3,$ prove $\frac{x^2}{y^2+yz+z^2}+\frac{y^2}{x^2+xz+z^2}+\frac{z^2}{y^2+yx+x^2}\ge \frac{5}{3}(x^2+y^2+z^2)+2xyz-6.$

Question

Let $x,y,z>0: x+y+z=3.$ Prove that $$\frac{x^2}{y^2+yz+z^2}+\frac{y^2}{x^2+xz+z^2}+\frac{z^2}{y^2+yx+x^2}\ge \frac{5}{3}(x^2+y^2+z^2)+2xyz-6.$$ I tried to prove the well-known result $$\frac{x^2}{y^2+yz+z^2}+\frac{y^2}{x^2+xz+z^2}+\frac{z^2}{y^2+yx+x^2}\ge 1.$$By applying Cauchy-Schwarz, $$\sum_{cyc}\frac{x^2}{y^2+yz+z^2}\ge \frac{(x^2+y^2+z^2)^2}{2(x^2y^2+y^2z^2+z^2x^2)+xyz(x+y+z)}\ge 1$$ $$\iff x^4+y^4+z^4\ge xyz(x+y+z),$$ which is true since $$x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xyz(x+y+z)$$ The rest is proving $$7\ge \frac{5}{3}(x^2+y^2+z^2)+2xyz$$I check $a=b=3/2;c=0$ and see that it is not true.

I hope to see some good ideas. It would be useful if you post a proof with the motivation.

Thank you for interest.

Answer 1

We need to prove that: $$\sum_{cyc}\frac{x^2}{y^2+yz+z^2}\geq\frac{15(x^2+y^2+z^2)}{(x+y+z)^2}+\frac{54xyz}{(x+y+z)^3}-6.$$ Now, let $x+y+z=3u, xy+xz+yz=3v^2$ and $xyz=w^3$.

Thus, we need to prove that: $$\frac{\sum\limits_{cyc}x^2(x^2+xy+y^2)(x^2+xz+z^2)}{\prod_{cyc}(x^2+xy+y^2)}\geq\frac{5(3u^2-2v^2)}{u^2}+\frac{2w^3}{u^3}-6$$ or $$\frac{27u^6-45u^4v^2+18u^2v^4-2v^6+4u^3w^3-uv^2w^3}{3uv^4-u^3w^3-v^6}\geq\frac{5(3u^2-2v^2)}{u^2}+\frac{2w^3}{u^3}-6$$ of $f(w^3)\geq0,$ where $$f(w^3)=2u^3w^6+(13u^6-11u^4v^2-6u^2v^4+2v^6)w^3+27u^9-45u^7v^2-9u^5v^4+37u^3v^6-10uv^8.$$ But by Schur $$f'(w^3)=4u^3w^3+13u^6-11u^4v^2-6u^2v^4+2v^6\geq$$ $$\geq4u^3(4uv^2-3u^3)+13u^6-11u^4v^2-6u^2v^4+2v^6=$$ $$=u^6+5u^4v^2-6u^2v^4+2v^6\geq0,$$ which says that $f$ increases.

Therefore, by $uvw$ it's enough to prove our inequality in the following cases.

1. $w^3=0$.

Let $z=0$ and $y=1$.

Thus, we need to prove that: $$x^2+\frac{1}{x^2}\geq\frac{15(x^2+1)}{(x+1)^2}-6$$ or $$x^6+2x^5-8x^4+12x^3-8x^2+2x+1\geq0,$$ which after substitution $x^2+1=2tx,$ where $t\geq1$, gives: $$8t^3-6t+2(4t^2-2)-16t+12\geq0$$ or $$4t^3+4t^2-11t+4\geq0$$ or $$(2t-1)(2t^2+3t-4)\geq0,$$ which is true for $t\geq1$.

2. Two variables are equal.

Let $y=z=1$.

Thus, we need to prove that: $$\frac{x^2}{3}+\frac{2}{x^2+x+1}\geq\frac{15(x^2+2)}{(x+2)^2}+\frac{54x}{(x+2)^3}-6$$ or $$(x-1)^2(x^5+9x^4+9x^3+26x^2+24x+12)\geq0$$ and we are done.

About $uvw$ see here: https://artofproblemsolving.com/community/c6h278791

Answer 2

Note that the original inequality is equivalent to $$ \frac{1}{25245}\sum \left(a - b\right)^{2} \cdot \left(623 a^{2} b \left(a - b\right)^{2} \left(b - c\right)^{2} + 4479 a^{2} c \left(a - c\right)^{2} \left(a - b\right)^{2} + 23080 a b^{4} \left(b - c\right)^{2} + 178 a b^{2} \left(a - c\right)^{2} \left(a - b\right)^{2} + 1692 a b c \left(a - b\right)^{4} + 14180 b^{4} c \left(b - c\right)^{2} + 11154 b^{2} c \left(a - b\right)^{2} \left(b - c\right)^{2} + 19242 c^{5} \left(a - b\right)^{2}\right) + \frac{1}{25245}\sum \left(a^{3} - b c^{2}\right)^{2} \cdot \left(18185 a^{3} + 73421 a^{2} c + 123999 a b c + 4005 a c^{2} + 6505 b^{2} c\right) + \frac{1}{25245}\sum \left(a^{3} - b^{2} c\right)^{2} \cdot \left(7060 a^{3} + 75646 a^{2} b + 110649 a b c + 10955 b c^{2}\right) + \frac{2413}{2805}\sum \left(a^{5} b^{4} + a^{4} b^{5} - 3 a^{4} b^{4} c + a^{3} b^{3} c^{3}\right) \ge 0 $$

Answer 3

By using AM-GM $$\frac{3x^2}{y^2+yz+z^2}+2.x^2\sqrt{\frac{y^2+yz+z^2}{3}}\ge 3x^2,$$and $$\sqrt{\frac{y^2+yz+z^2}{3}}=\sqrt{\frac{y+z+\sqrt{yz}}{3}(y+z-\sqrt{yz})}\le \frac{1}{2}\left(\frac{y+z+\sqrt{yz}}{3}+y+z-\sqrt{yz}\right).$$Thus $$\frac{x^2}{y^2+yz+z^2}\ge x^2+\frac{2x\sqrt{yz}-4(xy+xz)}{9}. \tag{1}$$ Take cyclic sum on $(1),$ we obtain $$\sum_{cyc}\frac{x^2}{y^2+yz+z^2}\ge x^2+y^2+z^2+\frac{2}{9}(z\sqrt{xy}+x\sqrt{yz}+y\sqrt{zx})-\frac{8}{9}(xy+yz+zx).$$ Id est, it's enough to prove that $$x^2+y^2+z^2+\frac{2}{9}(z\sqrt{xy}+x\sqrt{yz}+y\sqrt{zx})-\frac{8}{9}(xy+yz+zx)\ge \frac{5}{3}(x^2+y^2+z^2)+2xyz-6,$$or$$z\sqrt{xy}+x\sqrt{yz}+y\sqrt{zx}+2(xy+yz+zx)\ge 9 xyz,$$which is true by $0<xyz\le 1$ and AM-GM $$z\sqrt{xy}+x\sqrt{yz}+y\sqrt{zx}+xy+yz+zx+xy+yz+zx\ge 9\sqrt[9]{(xyz)^6}\ge 9xyz.$$ The proof is done. Equality holds at $x=y=z=1.$