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If $a$, $b$, $c$ are positive real numbers such that $a+b+c = 1$, prove that $$(1+a)(1+b)(1+c) ≥ 8(1-a)(1-b)(1-c)\text{.}$$

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    $\begingroup$ The question doesn't show any effort. $\endgroup$ – Apurv Feb 8 '14 at 10:15
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One quite natural start is to replace $1-a$ by $b+c$ as there are more inequalities for nonnegative reals than reals. (In this case all terms are positive but minus signs are ugly in this problem so let's remove them.) So we have to prove that $$(1+a)(1+b)(1+c)\geq 8(b+c)(a+c)(a+b).$$ Also, there are number eight in the inequality so that makes me guess that one might be able to prove the inequality by taking three times arithmetic-geometric inequality. So I have to modify terms somehow. Lets test: $$1+a=a+b+c+a=2a+b+c,1+b=2b+a+c,1+c=2c+a+b.$$ But arithmetic-geometric inequality means that we have to take square root of expressions so squares would be nice. There are something similar terms in $2a+b+c$ and $2b+a+c,$ namely $a+b$. So lets try to write the inequality as $$((a+b)+(a+c))((a+b)+(b+c))((a+c)+(b+c))\geq 8(b+c)(a+c)(a+b).$$ But this is just the arithmetic-geometric inequality of the numbers $a+b,a+c,$ and $b+c.$

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$a+b, c+a$ is positive real, consider $\frac{1+a}{2}= \frac{(a+b)+(c+a)}{2} \geq \sqrt{(a+b)(c+a)}$ i.e$(1+a)\geq2\cdot\sqrt{(1-c)(1-b)}$ and so..

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Note that by A.M - G.M we have $$ (b+c)+(c+a) \geq 2 \cdot \sqrt{(b+c)\cdot(c+a)}$$ This says $2c+b+a =1+c\geq 2 \cdot \sqrt{(b+c)\cdot (c+a)}$. Similary get inequalities for $(1+b)$ and $(1+c)$ multiply and get the answer.

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We can easily prove the inequality: $(a+b)(b+c)(c+a)\ge 8abc$.

Replacing $a$ by $(1-a)$, $b$ by $(1-b)$ and $c$ by $(1-c)$ we get:

$$(2-(a+b))(2-(c+b))(2-(c+a))\geq 8(1-a)(1-b)(1-c)$$

but $a + b + c= 1$ therefore:

$$a+b=1-c,~ b+c=1-a, ~\text{and}~ c+a=1-b$$

substituting these values in LHS we get :

$$(1+a)(1+b)(1+c)\ge8(1−a)(1−b)(1−c)$$

which proves the inequality.

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