For $x$, $y$, $z$ the sides of a triangle, show $\sum_{cyc}\frac{yz((y+z)^2-x^2)}{(y^2+z^2)^2}\ge\frac{9(y+z-x)(x+z-y)(x+y-z)}{4xyz}$

Question

in $\triangle ABC$, let $AB=z,BC=x,AC=y$,show that $$\sum_{cyc}\frac{yz((y+z)^2-x^2)}{(y^2+z^2)^2}\ge\frac{9(y+z-x)(x+z-y)(x+y-z)}{4xyz}$$

by well kown Iran 96 inequality $$(xy+yz+xz)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\dfrac{1}{(x+z)^2}\right)\ge\dfrac{9}{4}$$

Answer 1

We need to prove that: $$\sum_{cyc}\frac{yz}{(y^2+z^2)^2(x+y-z)(x+z-y)}\geq\frac{9}{4xyz(x+y+z)}.$$ Now, by C-S $$\sum_{cyc}\frac{yz}{(y^2+z^2)^2(x+y-z)(x+z-y)}\geq$$ $$\geq\sum_{cyc}\frac{y^2z^2}{yz(y^2+z^2)^2(x+y-z)(x+z-y)}\geq$$ $$\geq\frac{(xy+xz+yz)^2}{\sum\limits_{cyc}yz(y^2+z^2)^2(x+y-z)(x+z-y)}.$$ Thus, it's enough to prove that $$4xyz(x+y+z)(xy+xz+yz)^2\geq9\sum\limits_{cyc}yz(y^2+z^2)^2(x+y-z)(x+z-y)$$ or $$\sum_{sym}(9x^7y-18x^6y^2+27x^5y^3-18x^4y^4-9x^5y^2z+4x^4y^3z+4x^4y^2z^2+x^3y^3z^2)\geq0$$ or $$9\sum_{cyc}xy(x^6-2x^5y+3x^4y^2-4x^3y^3+3x^2y^4-2xy^5+y^6)-$$ $$-9xyz\sum_{cyc}xy(x^3-x^2y-xy^2+y^3)-xyz\sum_{cyc}(5x^3y^2+5x^3z^2-8x^3yz-2x^2y^2z)\geq0$$ or $$\sum_{cyc}(x-y)^2xy(9(x^2+y^2)^2-9xyz(x+y)-5z^4-xyz^2)\geq0$$ or $$7\sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)+$$ $$+\sum_{cyc}(x-y)^2xy(2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4))\geq0.$$

We'll prove that $$\sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)\geq0.$$ Indeed, let $x\geq y\geq z$.

Thus, $$\sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)\geq$$ $$\geq(x-z)^2xz(x^4+z^4-y^4)+(y-z)^2yz(y^4+z^4-x^4)\geq$$ $$\geq(y-z)^2xz(x^4-y^4)+(y-z)^2yz(y^4-x^4)=$$ $$=z(y-z)^2(x^4-y^4)(x-y)\geq0.$$ Thus, it's enough to prove that: $$2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4)\geq0.$$ We'll prove that: $$\sqrt[4]{\frac{x^4+9x^2y^2+y^4}{11}}\geq\sqrt[3]{\frac{xy(x+y)}{2}}.$$ Let $x^2+y^2=2uxy$.

Thus, $u\geq1$ and we need to prove that: $$\left(\frac{x^4+9x^2y^2+y^4}{11}\right)^3\geq\left(\frac{xy(x+y)}{2}\right)^4$$ or $$\left(\frac{4u^2-2+9}{11}\right)^3\geq\frac{(2u+2)^2}{16}$$ or $f(u)\geq0,$ where $$f(u)=3\ln(4u^2+7)-2\ln(u+1)+2\ln2-3\ln11.$$ But $$f'(u)=\frac{24u}{4u^2+7}-\frac{2}{u+1}>0,$$ which says $f(u)\geq f(1)=0.$

Also, by AM-GM $$\sqrt[3]{\frac{xy(x+y)}{2}}\geq\sqrt{xy}.$$

Now, let $z=t\sqrt[3]{\frac{xy(x+y)}{2}}.$

Thus, $$2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4)\geq$$ $$\geq2z^4-\left(\sqrt[3]{\frac{xy(x+y)}{2}}\right)^2z^2-18\left(\sqrt[3]{\frac{xy(x+y)}{2}}\right)^3z+22\left(\sqrt[3]{\frac{xy(x+y)}{2}}\right)^4=$$ $$=\left(\sqrt[3]{\frac{xy(x+y)}{2}}\right)^4(2t^4-t^2-18t+22)\geq0.$$ Can you end it now?