Question:
let $x,y,z>0$, show that $$\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge 3+\dfrac{x^4+y^4+z^4-x^2-y^2-z^2}{x^3+y^3+z^3-xyz}$$
I know this well know inequality $$\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge \dfrac{37(x^2+y^2+z^2)-19(x+y+z)}{6(x+y+z)}$$
I found this inequality is stronger, and I can't prove it
BW helps.
We need to prove that $$\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\right)(x^3+y^3+z^3-xyz)-x^4-y^4-z^4+x^2+y^2+z^2\geq$$ $$\geq3(x^3+y^3+z^3-xyz).$$ Now, by AM-GM $$\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\right)(x^3+y^3+z^3-xyz)-x^4-y^4-z^4+x^2+y^2+z^2\geq$$ $$\geq2\sqrt{\left(\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\right)(x^3+y^3+z^3-xyz)-x^4-y^4-z^4\right)(x^2+y^2+z^2)}.$$ Thus, it's enough to prove that $$4\left(\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\right)(x^3+y^3+z^3-xyz)-x^4-y^4-z^4\right)(x^2+y^2+z^2)\geq$$ $$\geq9(x^3+y^3+z^3-xyz)^2.$$ Let $x=\min\{x,y,z\}$, $y=x+u$, $z=x+v$ and $u=kv$.
Hence, $$4xyz\left(\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\right)(x^3+y^3+z^3-xyz)-x^4-y^4-z^4\right)(x^2+y^2+z^2)-$$ $$-9xyz(x^3+y^3+z^3-xyz)^2=$$ $$=8(u^2-uv+v^2)x^7-8uv(u-8v)x^6+3(5u^4-34u^3v+71u^2v^2+14uv^3+5v^4)x^5+$$ $$+(21u^5-125u^4v+201u^3v^2+121u^2v^3+95uv^4+21v^5)x^4+$$ $$+(29u^6-141u^5v+238u^4v^2-29u^3v^3+198u^2v^4+51uv^5+29v^6)x^3+$$ $$+(19u^7-75u^6v+114u^5v^2+12u^4v^3+52u^3v^4+90u^2v^5+29uv^6+19v^7)x^2+$$ $$+(4u^8-13u^7v+16u^6v^2+24u^5v^3-2u^4v^4+40u^3v^5+8u^2v^6+19uv^7+4u^8)x+$$ $$+4uv^3(u^5+u^3v^2+u^2v^3+v^5)\geq$$ $$=8(u^2-uv+v^2)x^7-8uv(u-8v)x^6+3(5u^4-34u^3v+71u^2v^2+14uv^3+5v^4)x^5\geq$$ $$\geq2\sqrt{24x^{12}(u^2-uv+v^2)(5u^4-34u^3v+71u^2v^2+14uv^3+5v^4)}-8uv(u-8v)x^6.$$ Id est, it remains to prove that $$3(k^2-k+1)(5k^4-34k^3+71k^2+14k+5)\geq2k^2(k-8)^2$$ or $$15k^6-117k^5+328k^4-241k^3+58k^2+27k+15\geq0,$$ which is true: $$15k^6-117k^5+328k^4-241k^3+58k^2+27k+15>$$ $$>k^2(15k^4-117k^3+328k^2-241k+58)=$$ $$=15\left(k^2-\frac{117}{30}k+\frac{3}{2}\right)^2+\frac{1}{20}(1097k^2-1310k+485)>0$$ and we are done!