a,b,c,d are real ,prove this

ADARSH · #1 $Old$ December 2nd, 2008, 12:36 AM

Given a,b,c,d as 4 real numbers

Avoiding method of contradiction prove

$a-b^2 , b - c^2 , c-d^2 , d- a^2$
cant all be greater
than $\frac{1}{4}$ ?
Thanks to the helper

flyingsquirrel · #2 $Old$ December 2nd, 2008, 02:17 AM

Hi,

Quote:

Originally Posted by ADARSH $View Post$

Given a,b,c,d as 4 real numbers

Avoiding method of contradiction prove

$a-b^2 , b - c^2 , c-d^2 , d- a^2$
cant all be greater
than $\frac{1}{4}$ ?

Let $f(a,b,c,d)=(a-b^2)+(b-c^2)+(c-d^2)+(d-a^2)$ .

$\begin{aligned}f(a,b,c,d)&=(a-b^2)+(b-c^2)+(c-d^2)+(d-a^2)\\ &=1{\color{blue}-a^2+a-\frac14}{\color{green}-b^2+b-\frac14}{\color{red}-c^2+c-\frac14}-d^2+d-\frac14\\ &=1\underbrace{{\color{blue}-\left(a-\frac12\right)^2}{\color{green}-\left(b-\frac12\right)^2}{\color{red}-\left(c-\frac12\right)^2}-\left(d-\frac12\right)^2}_{\leq 0}\\ f(a,b,c,d)&\leq 1\\ \end{aligned}$

Can you take it from here ?