If $a^2+b^2+c^2+abc=4$, Find minimum $P=\sqrt{\frac{2a+bc}{3}}+\sqrt{\frac{2b+ca}{3}}+\sqrt{\frac{2c+ab}{3}}-\frac{3(a+b+c+abc)}{2}$

Question

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

Answer 1

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.

Answer 2

The minimum is $-3$ when $a = b = c = 1$.

Sketch of a proof.

We have, for all $x \ge 0$, $$\sqrt x \ge -\frac12x^2 + \frac32 x. \tag{1}$$

Using (1), it suffices to prove that $$\sum_{\mathrm{cyc}} \left[-\frac12\left(\frac{2a+bc}{3}\right)^2+ \frac32 \left(\frac{2a+bc}{3}\right)\right] - \frac{3(a + b + c + abc)}{2} \ge -3$$ or \begin{align*} &54 + 9(ab + bc + ca) - 4(a^2 + b^2 + c^2) - (a^2b^2 + b^2c^2 + c^2a^2) \\ &\quad- 39abc - 9(a + b + c) \ge 0. \tag{2} \end{align*}

We can prove (2) by the pqr method.

Let $p = a + b + c, q = ab + bc + ca, r = abc$.

The condition $a^2 + b^2 + c^2 + abc = 4$ is written as $$p^2 - 2q + r = 4. \tag{3}$$

From (3), we have $$2 \le p \le 3, \quad 0\le q \le 3p - \sqrt{3p(p + 1)(4 - p)}. \tag{4}$$

From (3), using $r = 4 + 2q - p^2$, (2) is written as $$-(q + 61/2 - 2p)^2 - 2p^3 + 39p^2 - 123p + \frac{3313}{4} \ge 0. \tag{5}$$

From (4) and (5), it suffices to prove that, for all $2 \le p \le 3$, $$-\left(3p - \sqrt{3p(p + 1)(4 - p)} + 61/2 - 2p\right)^2 - 2p^3 + 39p^2 - 123p + \frac{3313}{4} \ge 0$$ which is true.

We are done.