inequality 2

great_math · #1 $Old$ 10-08-2008, 10:15 PM

If a,b and c are non-negative reals such that $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \ge 2$ , prove that

$abc \le \frac{1}{8}$

thank you.

NonCommAlg · #2 $Old$ 10-09-2008, 09:31 PM

Quote:

Originally Posted by great_math $View Post$

If a,b and c are non-negative reals such that $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \ge 2.$ prove that $abc \le \frac{1}{8}.$

let $a=\tan^2x, \ b = \tan^2 y, \ c=\tan^2 z, \ \ 0 \leq x,y,z < \frac{\pi}{2}.$ so your inequality becomes: $\cos^2x + \cos^2y+\cos^2z \ge 2.$ hence:

$\boxed{1} \ \ \cos^2x \ge 2 - \cos^2y - \cos^2z = \sin^2y + \sin^2z \ge 2\sin y \sin z.$ similarly:

$\boxed{2} \ \ \cos^2 y \ge 2\sin x \sin z,$

$\boxed{3} \ \ \cos^2 z \ge 2 \sin x \sin y.$

multiplying $\boxed{1} \ , \boxed{2} \ , \boxed{3} \ ,$ together will give us: $\cos^2x \cos^2y \cos^2z \ge 8 \sin^2 x \sin^2y \sin^2 z.$ thus: $\tan^2x \tan^2y \tan^2z \le \frac{1}{8}. \ \ \ \Box$

NonCommAlg · #3 $Old$ 10-12-2008, 12:11 PM

for those who are interested, the above inequality can be easily generalized, i.e. if $a_1, \cdots , a_n, \ n \geq 2,$ are non-negative numbers and $\frac{1}{1+a_1} + \frac{1}{1+a_2}+\cdots + \frac{1}{1+a_n} \geq n-1,$ then:

$a_1a_2 \cdots a_n \leq \frac{1}{(n-1)^n}.$ the proof is the same and is based on AM-GM: let $a_j=\tan^2x_j, \ 0 \leq x_j < \frac{\pi}{2}, \ 1 \leq j \leq n.$ then the inequality becomes: $\cos^2x_1 + \cos^2x_2 + \cdots + \cos^2x_n \geq n-1.$ so:

$\cos^2 x_j \geq n-1 - \sum_{i \neq j} \cos^2x_i=\sum_{i \neq j} \sin^2 x_i \geq (n-1) \left(\prod_{i \neq j} \sin^2x_i \right)^{\frac{1}{n-1}}, \ \ 1 \leq j \leq n.$ multiplying these inequalities together gives us: $\cos^2 x_1 \cos^2 x_2 \cdots \cos^2x_n \geq (n-1)^n \sin^2x_1 \sin^2x_2 \cdots \sin^2x_n.$

hence: $a_1 a_2 \cdots a_n = \tan^2x_1 \tan^2x_2 \cdots \tan^2x_n \leq \frac{1}{(n-1)^n}. \ \ \ \Box$