Inequality. (1 Viewer)

[seanieg89]

Let a,b,c > 0 with abc=1. Prove that:

 

[Fus Ro Dah]

The solution falls out fairly quickly if you use the following identities, but I'm just doing this over the top of my head at the moment. I'll look for a quicker method when I have the time ^^

 

[abecina]

 

[abecina]

I am sure that says something about the lack of elegance in my solution. Here it is without the tex

\begin{align*}
abc &= 1
\Rightarrow ((a+1)-1)((b+1)-1)((c+1)-1) &= 1\\
(a+1)(b+1)(c+1) - [(a+1)(b+1)+(a+1)(c+1)+(b+1)(c+1)] + [(a+1)+(b+1)+(c+1)] -1 &= 1\\
\textrm{Dividing through by }(a+1)(b+1)(c+1) \textrm{ and rearranging gives}\\
\left(\frac{1}{(b+1)(c+1)} - \frac{1}{c+1}\right)+\left(\frac{1}{(a+1)(c+1)}- \frac{1}{b+1}\right)+\left(\frac{1}{(a+1)(b+1)}- \frac{1}{a+1} \right) &= \frac{2}{(a+1)(b+1)(c+1)}-1\\
-\left(\frac{b}{(b+1)(c+1)}+\frac{a}{(a+1)(b+1)}+\frac{c}{(c+1)(a+1)}\right) &= \frac{2}{(a+1)(b+1)(c+1)} - 1\\
\left(\frac{b}{(b+1)(c+1)}+\frac{a}{(a+1)(b+1)}+\frac{c}{(c+1)(a+1)}\right) &= 1- \frac{2}{(a+1)(b+1)(c+1)}\\
\textrm{ From the AM-GM inequality} a+1&\geq 2\sqrt{a}\\
b+1&\geq 2\sqrt{b}\\
c+1&\geq 2\sqrt{c}\\
\Rightarrow (a+1)(b+1)(c+1)&\geq 8\sqrt{abc} = 8\\
\Rightarrow \frac{1}{(a+1)(b+1)(c+1)} &\leq \frac{1}{8}\\
\Rightarrow -\frac{2}{(a+1)(b+1)(c+1)}&\geq -\frac{1}{4}\\
\Rightarrow 1 - \frac{2}{(a+1)(b+1)(c+1)}&\geq 1 - \frac{1}{4} = \frac{3}{4}\\
\therefore \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)} &\geq \frac{3}{4}
\end{align*}

 

[Shadowdude]

what might work is if you put 'tex' tags on every line


But... I don't know how that'll look.

 

[hamster-lol]

I can't wait to do inequalities in 4u! That looks so cool!

 

[abecina]

 

[Fus Ro Dah]

View attachment 26515

Make the denominator of the entire term become (a+1)(b+1)(c+1).