Inequality problem

**Mathroum** · February 11th 2008, 02:10 PM

Could anyone help me solving this inequality?

a,b,c are positive real numbers, proove that:

[a/(b+c) + 1/2]*[b/(a+c) + 1/2]*[c/(a+b) + 1/2] >- +1

Thank You

I tried to do it with LaTEx:

$( \frac{a}{b+c} + \frac{1}{2})*( \frac{b}{a+c} + \frac{1}{2})*( \frac{c}{a+b} + \frac{1}{2}) \geq 1$

OR

$[ \frac{a}{b+c} + \frac{1}{2}][ \frac{b}{a+c} + \frac{1}{2}][ \frac{c}{a+b} + \frac{1}{2}] \geq 1$

**Mathroum** · February 12th 2008, 09:18 AM

Hey I succeeded in prooving the inequality. The solution :

$\frac{(2a+b+c)(2b+a+c)(2c+a+b)}{8(b+c)(a+c)(a+b)} \geq 1$

when we multiply the whole inequality by $8(b+c)(a+c)(a+b)$ and get rid of brackets, we get :

$2a^3+2b^3+2c^3-a^2b-ab^2-b^2c-a^2c-ac^2-bc^2 \geq 0$

$a^3+b^3-a^2b-ab^2 = (a+b)^3-4a^2b-4ab^2$
$a^3+c^3-a^2c-ac^2 = (a+c)^3-4a^2c-4ac^2$
$b^3+c^3-b^2c-bc^2 = (b+c)^3-4b^2c-4bc^2$

then :

$(a+b)^3-4ab(a+b) = (a+b)[(a+b)^2 - 4ab] = (a+b)(a-b)^2$
... same for (a+c) and (b+c)....

Finally : $(a+b) \geq 0 , (a-b)^2 \geq 0 \Rightarrow (a+b)(a-b)^2 \geq 0$

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