# Inequality

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• February 20th 2008, 08:21 AM
james_bond
Inequality
Prove that if $ab + ac + bc = 3$:
$8(a + b + c)^2\geq9(a + b)(a + c)(b + c)$
( $a,b,c\ge 0$)
• February 21st 2008, 06:17 AM
wingless
The question says $a,b,c\ge 0$..
• February 22nd 2008, 11:35 AM
JaneBennet
Suppose one of $a,b,c$ is 0. Then the other two can’t be 0 if the condition $ab+bc+ca=3$ is to hold. Say, we have $a=0$. Then $bc=3$ and the inequality becomes

$\color{white}.\quad.$ $8\left(b+\frac{3}{b}\right)^2\geq9bc\left(b+\frac{ 3}{b}\right)=27\left(b+\frac{3}{b}\right)$

$\Leftrightarrow\ 8\left(b+\frac{3}{b}\right)\geq27$

$\Leftrightarrow\ 8b^2-27b+24\geq0$

The LHS, as a quadratic expression in $b$, has negative discriminant and so the inequality is true.

So now we’ll assume $a,b,c>0$. Let $a=\sqrt{3}\,A,\ b=\sqrt{3}\,B,\ c=\sqrt{3}\,C$. Then $AB+BC+CA=1$ and the inequality becomes

$8(A+B+C)^2\geq9\sqrt{3}(A+B)(B+C)(C+A)$

Since $A,B,C$ are positive and $AB+BC+CA=1$, we can make the following substitutions:

$A=\tan{\alpha},\ B=\tan{\beta},\ C=\tan{\gamma}$

where $\alpha+\beta+\gamma=\frac{\pi}{2}$. Then observe that

$A+B=\frac{\sin{\alpha}}{\cos{\alpha}}+\frac{\sin{\ beta}}{\cos{\beta}}$

$\color{white}.\hspace{10mm}.$ $=\frac{\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\b eta}}{\cos{\alpha}\cos{\beta}}$

$\color{white}.\hspace{10mm}.$ $=\frac{\sin{\left(\alpha+\beta\right)}}{\cos{\alph a}\cos{\beta}}$

$\color{white}.\hspace{10mm}.$ $=\frac{\cos{\left(\frac{\pi}{2}-\left[\alpha+\beta\right]\right)}}{\cos{\alpha}\cos{\beta}}$

$\color{white}.\hspace{10mm}.$ $=\frac{\cos{\gamma}}{\cos{\alpha}\cos{\beta}}$

Hence

$(A+B)(B+C)(C+A)$ $=$ $\left(\frac{\cos{\gamma}}{\cos{\alpha}\cos{\beta}} \right)\left(\frac{\cos{\alpha}}{\cos{\beta}\cos{\ gamma}}\right)\left(\frac{\cos{\beta}}{\cos{\gamma }\cos{\alpha}}\right)=\frac{1}{\cos{\alpha}\cos{\b eta}\cos{\gamma}}$

So it comes down to proving this inequality:

$\cos{\alpha}\cos{\beta}\cos{\gamma}\left(\tan{\alp ha}+\tan{\beta}+\tan{\gamma}\right)^2\ \geq\ \frac{9\sqrt{3}}{8}$
• February 22nd 2008, 08:21 PM
Charbel
hmmm... don’t know why I was heavily neg repped on this question =\ but yer .. anyways...

^^ top was interesting

After about 2 pages of working trying to find similarities between the two I got this...

LHS

8(a+b+c)^2

8a^2 + 16ab + 16ac + 8b^2 + 16bc + 8c^2

8(a^2 + b^2 + c^2) + 16(ac + ab + cd)

8(a^2 + b^2 + c^2) + 16(3)

8(a^2 + b^2 + c^2) + 48

8(a^2 + b^2 + c^2 + 6)

RHS

9(a+b)(a+c)(b+c)

9a^2b + 9a^2c + 9ab^2 + 18abc + 9ac^2 + 9b^2c + 9bc^2

9a(ab + ac) + 9b(ab + bc) + 9c(ac + bc) +18(abc)

9a(3-bc) + 9b(3-ac) + 9c(3-ab) +18abc

27a – 9abc + 27b – 9abc + 27c -9abc + 18abc

27(a + b + c) – 9abc

9(3(a + b + c) - abc)

So we can say….

8(a^2 + b^2 + c^2 + 6) => 9(3(a + b + c) - abc)

when ab + cb + bc = 3…..

I also noticed a + b + c appears In the RHS which also appears in the LHS

So we can let a + b + c = d

And say:

8d^2 => 27d – 9abc

still doesn’t help unless we know a b or c = 0…