Show that for all positive real numbers $a$, $b$ and $c$ such that $abc=1$, the inequality $a+b+c+2a^4+2b^4+2c^4\ge \dfrac{3}{2}\left(a^2\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b^2\left(\dfrac{1}{a}+\dfrac{1}{c}\right)+c^2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right)$ is true
I tried using Schur inequality which gave me $a+b+c+a^4+b^4+c^4\ge a^2\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b^2\left(\dfrac{1}{a}+\dfrac{1}{c}\right)+c^2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)$. Then I wanted to show that $a^4+b^4+c^4\ge\dfrac{1}{2}\left(a^2\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b^2\left(\dfrac{1}{a}+\dfrac{1}{c}\right)+c^2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right)$ but I'm not sure how to go about that
We need to prove that $$\sum_{cyc}(2a^4+a^2bc)\geq\frac{3}{2}\sum_{cyc}a^3(b+c)$$ or $$\sum_{cyc}(4a^4-3a^3b-3a^3c+2a^2bc)\geq0$$ or $$3\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+\sum_{cyc}(a^4-a^2bc)\geq0,$$ which is true by Schur and Muirhead.