$a+b+c=1$, prove $(1+a)(1+b)(1+c) \ge 8(1-a)(1-b)(1-c)$, without AM-GM

$$a,b,c \in \mathbb{R}^{+}$$, if $$a+b+c=1$$, prove that $$(1+a)(1+b)(1+c) \ge 8(1-a)(1-b)(1-c)$$

but without AM-GM as the only tool.

Collecting data:

Since $$a+b+c=1$$, we cannot have one of the variable to be greater than or equal to 1. So $$\boxed{0 < a,b,c < 1}$$ Also by AM-GM: $$1=a+b+c \ge 3(abc)^{1/3}$$, or

$$\boxed{abc \le \frac{1}{27}}$$

Next, $$(1-a)(1-b)(1-c) = [1 - (a+b) + ab](1-c) = \boxed{ 1 - (a+b+c) + (ab + ac + bc) - abc }$$ $$=\boxed{ (ab+ac+bc)-abc}$$

Next, $$(1+a)(1+b)(1+c) = [1 + (a+b) + ab](1+c) = 1 + (a+b+c) + (ac+ab+bc) + abc$$ $$= \boxed{2 + (ac+bc+bc) + abc}$$

so the inequality is equivalent with:

$$2 + ac+bc+ab + abc \ge 8(ab+ac+bc) - 8abc$$ or

$$\boxed{ 2 + 9abc \ge 7 (ab+ac+bc) } \:\: \leftarrow \:\: \text{to be proven}$$

Next, $$(a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab+ac+bc) = 1$$, $$\boxed{ab+ac+bc = \frac{1- (a^{2} + b^{2}+ c^{2})}{2}}$$

which means $$7(ab+ac+bc) = \frac{7}{2} -\frac{7}{2}(a^{2}+b^{2}+c^{2}) = 2 + \frac{3 - 7(a^{2}+b^{2}+c^{2})}{2}$$

and it is left to prove

$$\frac{3-7(a^{2}+b^{2}+c^{2})}{2} \le 9abc$$

$$\boxed{3 \le 18 abc + 7(a^{2}+b^{2}+c^{2})} \:\: \leftarrow \:\: \text{to be proven}$$

Hint:
As an alternative, note $$x \mapsto \log\dfrac{1+x}{1-x}$$ is convex, and use Jensen's inequality.

• maybe you mean $f(x) = \log ( \frac{1-x}{1+x} )$, it is also convex. Then we hope Jensens can show $\log ( \frac{1-a}{1+a} ) \le \frac{1}{2}$ also for $b,c$. Jul 13, 2022 at 3:32
• But how can we use it? involves $t=a, \: 1-t = b+c$? Jul 13, 2022 at 3:51
• Read up on Jensen's inequality from the link provided or any other source. Using Jensen's inequality, if $f$ is convex, then $f(a)+f(b)+f(c) \geqslant 3\cdot f\left(\frac{a+b+c}3\right)$. That gives you the answer directly. Jul 13, 2022 at 4:05
• That is not what I read in wiki, ok thanks. Jul 13, 2022 at 4:25
• Refer to the simplest "Finite Form" section - there is also an e.g. given where AM-GM is proved using Jensen. There are a lot of resources on this inequality, including online ones like brilliant.org/wiki/jensens-inequality Jul 13, 2022 at 6:37

$$a+b+c=1.$$ Prove that $$(1+a)(1+b)(1+c)\geq 8(1-a)(1-b)(1-c).$$

ETS) $$(2a+b+c)(a+2b+c)(a+b+2c) \geq 8(a+b)(b+c)(c+a).$$

Let $$b+c=x, c+a=y, a+b=z.$$

$$\Leftrightarrow (x+y)(y+z)(z+x) \geq 8xyz.$$

You may try from here.

Hint: $$(x+y)^2\geq4xy.$$

• I know this solution. But what if without AM-GM only? Jul 12, 2022 at 10:15
• I didn't use any AM-GM for this. $(x+y)^2 \geq 4xy$ can be induced by $(x-y)^2\geq0.$
– RDK
Jul 12, 2022 at 10:16
• If you don't like this solution, try to solve this with Muirhead inequality by multiplying this out.
– RDK
Jul 12, 2022 at 10:19
• But it can be considered as AM-GM. Thanks for the answer +1, but I am looking for other solutions. Jul 12, 2022 at 10:19
• Or, such a famous solution, with Jensen inequality. Set $f(x)=\ln \frac {1+x}{1-x},$ then try to use Jensen inequality.
– RDK
Jul 12, 2022 at 10:21

Proceeding along the OP:

It suffices to prove that $$2 + 9abc \ge 7(ab + bc + ca).$$

Degree three Schur inequality yields $$(a + b + c)^3 - 4(a + b + c)(ab + bc + ca) + 9abc \ge 0$$ that is $$1 - 4 (ab + bc + ca) + 9abc \ge 0.$$

It suffices to prove that $$2 + 4(ab + bc + ca) - 1 \ge 7(ab + bc + ca)$$ or $$1 \ge 3(ab + bc + ca)$$ which is true since $$(a + b + c)^2 - 3(ab + bc + ca) = a^2 + b^2 + c^2 - ab - bc - ca$$ $$= \frac12[(a-b)^2+(b-c)^2 + (c-a)^2]\ge 0$$.

We are done.