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

Question

$a,b,c \in \mathbb{R}^{+}$, 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.

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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

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

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

Answer 1

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

Answer 2

$a+b+c=1.$ 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: $(x+y)^2\geq4xy.$

Answer 3

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.