Given three real numbers $a,b,c$ so that $\{a, b, c\}\subset [1, 2]$ . Prove that $7abc\geq ab(a+ b)+ bc(b+ c)+ ca(c+ a)$ .

Question

I need to a fresh solution with $a:\neq {\rm mid}\{a, b, c\}$ , but mine

$$\begin{align*} 7abc &- ab(a+ b)- bc(b+ c)- ca(c+ a)= \\ &= a(2b- c)(2c- b)- (b+ c)(a- b)(a- c)\geq 0 \end{align*}$$

Answer 1

Setting $$P = 7abc - ab(a+ b)- bc(b+ c)- ca(c+ a).$$ Suppose $a \geqslant b \geqslant c,$ we have $$P= (2a-c)(2b-a)(2c-b)+3(a-b)(b-c)(a-c) \geqslant 0.$$ But I like $$2P= (2a-b)(2b-c)(2c-a)+(2a-c)(2b-a)(2c-b) \geqslant 0.$$

Answer 2

Let $$P = 7abc - ab(a+ b)- bc(b+ c)- ca(c+ a).$$

The second derivative in each variable is clearly $\leq 0$, so the minimum is achieved at the end points.

Checking them, the minimum is 0.