Let $a,b,c\ge 0: a^2+b^2+c^2+abc=4$. Find minimum $$P=\sqrt{\frac{2a+bc}{3}}+\sqrt{\frac{2b+ca}{3}}+\sqrt{\frac{2c+ab}{3}}-\frac{3(a+b+c+abc)}{2}$$
When $a=b=c=1,$ we get that $P\ge -3$ So we need to prove $$\sqrt{\frac{2a+bc}{3}}+\sqrt{\frac{2b+ca}{3}}+\sqrt{\frac{2c+ab}{3}}\ge \frac{3(a+b+c+abc)}{2}$$
I tried to use substitution $a=\dfrac{2x}{\sqrt{(x+y)(x+z)}}$ but it is complicated
Also, by AM-GM $$\sqrt{\frac{2a+bc}{3}}+\sqrt{\frac{2b+ca}{3}}+\sqrt{\frac{2c+ab}{3}}\ge 3\sqrt[3]{\sqrt{\frac{2a+bc}{3}}+\sqrt{\frac{2b+ca}{3}}\sqrt{\frac{2c+ab}{3}}}\ge\frac{3(a+b+c+abc-2)}{2}$$
But I don't know how to prove that last one.
Hope some user here can help. Thank you
Best Answer
Proof.
According to the given condition, we can demonstrate the identity$$\frac{a^3}{2a+bc}+\frac{b^3}{2b+ac}+\frac{c^3}{2c+ab}=2-abc.\tag{1}$$ (See here for detail proving.)
Now, apply AM-GM as$$\frac{3a^3}{2a+bc}+2.\sqrt{\frac{2a+bc}{3}}\ge 3a,$$and take cyclic sum on it, we obtain $$\sqrt{\frac{2a+bc}{3}}+\sqrt{\frac{2b+ca}{3}}+\sqrt{\frac{2c+ab}{3}}\ge \frac{3(a+b+c)}{2}-\frac{3}{2}.\sum_{cyc}\frac{a^3}{2a+bc}.\tag{2}$$ From $(1)$ and $(2),$ the desired result follows.