Let a , b , c possitive real numbers :
Show that :
a(b+c)2+b(c+a)2+c(a+b)2≤94(a+b+c)
341For those who are interested (and may their tribe increase!)
By Cauchy Schwarz Inequality
(a+b+c) \left(\frac{a}{(b+c)^2} + \frac{b}{(a+c)^2} + \frac{c}{(a+b)^2} \right)\ge \left(\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \right)^2
By Nesbitt Inequality
\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \ge \frac{3}{2}
Combining these two and rearranging we have the desired inequality
1Prophet, I thought the 3 main inequalities were sufficient. Oh well, another piece of knowledge..... :)
21WLOG a\geq b\geq c. \implies a+b\geq a+c\geq b+c
By chebyshevs inequality \frac{a}{(b+c)^2}+ \frac{b}{(c+a)^2}+ \frac{c}{(a+b)^2}\geq\frac{(a+b+c)}{3}\left ( \frac{1}{(b+c)^2} + \frac{1}{(c+a)^2}+ \frac{1}{(c+a)^2}\right )
Sufficient to prove \frac{(a+b+c)}{3}\left ( \frac{1}{(b+c)^2} + \frac{1}{(c+a)^2}+ \frac{1}{(c+a)^2}\right )\geq \frac{9}{4(a+b+c)}
With a+b= x,b+c=y,c+a=z(Known as Ravi's substitution)
This is equivalent to \left ( (x+y+z)(\frac{1}{x}+ \frac{1}{y} +\frac{1}{z}\right )^2\geq 81 \iff\left ( (x+y+z)(\frac{1}{x}+ \frac{1}{y} +\frac{1}{z}\right )\geq 9 Which is true.(by Am-Hm)
