Prove inequality

[baranka]

Let [MATH]1>a>0, 1>b>0, 1>c>0, a+b+c=1[/MATH]Prove that
[MATH]\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}.[/MATH]I saw the following solution. Let [MATH]x=\frac{2}{1-a^2}, y=\frac{2}{1-b^2}, z=\frac{2}{1-c^2}[/MATH], then, using AM-GM inequality, we get
[MATH]x+y+z-3\ge 3 \sqrt[3]{xyz}-3=\frac{27}{4}-3=\frac{15}{4}\; for \; x=y=z=\frac{9}{4}.[/MATH]Is it correct?

 

[MarkFL]

Hello, and welcome to FMH!

I agree with your work, and offer the method I would choose:

Using cyclic symmetry, the constraint tells us a critical value occurs for:

[MATH](a,b,c)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)[/MATH]
for which, if given:

[MATH]f(a,b,c)=\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}[/MATH]
Then:

[MATH]f\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)=\frac{15}{4}[/MATH]
Picking another point on the constraint, we find:

[MATH]f\left(\frac{1}{4},\frac{1}{4},\frac{1}{2}\right)=\frac{59}{15}>\frac{15}{4}[/MATH]
And so, we may conclude that:

[MATH]f_{\min}=\frac{15}{4}[/MATH]