Let $a,b$ and $c$ are positive numbers. Prove that $$\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}+15\sqrt[3]{abc}\geq 6(a+b+c).$$ I tried BW, SOS-Schur, but without any success.
I let $abc=1$. Then assuming $a=y/x,b=z/y,c=x/z$.
I got a six degree polynomial. But , I failed to complete the proof.
BW helps here!
We need to prove that $$\frac{a^6}{b^3}+\frac{b^6}{c^3}+\frac{c^6}{a^3}+15abc\geq6(a^3+b^3+c^3).$$ Now, let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Thus, we need to prove that: $$3(u^2-uv+v^2)a^{10}+(13u^3-45u^2v+36uv^2+13v^3)a^9+$$ $$+3(17u^4-55u^3v+18u^2v^2+26uv^3+17v^4)a^8+$$ $$+6(15u^5-35u^4v-12u^2v^2+24u^2v^3+28uv^4+15v^5)a^7+$$ $$+6(13u^6-18u^5v-30u^4v^2+13u^3v^3+33u^2v^4+45uv^5+13v^6)a^6+$$ $$+6(6u^7-3u^6v-18u^5v^2-5u^4v^3+16u^3v^4+45u^2v^5+39uv^6+6v^7)a^5+$$ $$+3(3u^8-6u^6v^2-12u^5v^3+5u^4v^4+30u^3v^5+78u^2v^6+36uv^7+3u^8)a^4+$$ $$+(u^9-6u^6v^3+78u^3v^6+108u^2v^7+27uv^8+v^9)a^3+$$ $$+3(12u^2+9uv+v^2)uv^7a^2+3u^2v^8a(3u+v)+u^3v^9\geq0,$$ which is true because $$3(u^2-uv+v^2)\geq3uv,$$ $$13u^3-45u^2v+36uv^2+13v^3\geq3\sqrt{u^3v^3},$$ $$3(17u^4-55u^3v+18u^2v^2+26uv^3+17v^4)\geq-21u^2v^2,$$ $$6(15u^5-35u^4v-12u^2v^2+24u^2v^3+28uv^4+15v^5)\geq-16\sqrt{u^5v^5},$$ $$6(13u^6-18u^5v-30u^4v^2+13u^3v^3+33u^2v^4+45uv^5+13v^6)\geq65u^3v^3,$$ $$6(6u^7-3u^6v-18u^5v^2-5u^4v^3+16u^3v^4+45u^2v^5+39uv^6+6v^7)\geq144\sqrt{u^7v^7},$$ $$3(3u^8-6u^6v^2-12u^5v^3+5u^4v^4+30u^3v^5+78u^2v^6+36uv^7+3u^8)\geq127u^4v^4$$ and $$u^9-6u^6v^3+78u^3v^6+108u^2v^7+27uv^8+v^9\geq88\sqrt{u^9v^9}.$$ Now, let $\frac{a}{\sqrt{uv}}=x$.
Thus, it's enough to prove that $$3x^7+3x^6-21x^5-16x^4+65x^3+144x^2+127x+88\geq0,$$ which is obvious.