1k good b555 , i followed the first method
Given a,b,c, are positive reals with abc=1
Prove that
1/a^3(b+c) + 1/b^3(a+c) + 1/c^3(a+b) \geq 3/2
-
UP 0 DOWN 0 0 6
6 Answers
39ok , there are 2 solns
1] take x=1/a , y=1/b , z=1/c
and now get the inequality in terms of x,y,z and use our CAUCHY SCHWARZ and then AM-GM INEQUALITY
2] i want to see if anyone does this............
USE REARRANGEMENT INEUALITY WHICH BHATT SIR INTRODUCED RECENTLY IN HIS QUESTION
1LHS\ becomes\ becomes\\ \frac{x^2}{(y+z)}+ \frac{y^2}{(x+z)}+ \frac{z^2}{(y+x)}\\ by\ rearrangement\ inequality\\ \frac{x^2}{(y+z)}+ \frac{y^2}{(x+z)}+ \frac{z^2}{(y+x)}\geq \frac{x^2}{(y+x)}+ \frac{y^2}{(y+z)}+ \frac{z^2}{(z+x)}\\ and\ similarly\\ \frac{x^2}{(y+z)}+ \frac{y^2}{(x+z)}+ \frac{z^2}{(y+x)}\geq \frac{z^2}{(y+z)}+ \frac{x^2}{(x+z)}+ \frac{y^2}{(y+x)}\\ adding\ both\ \frac{x^2}{(y+z)}+ \frac{y^2}{(x+z)}+ \frac{z^2}{(y+x)}\geq \frac{y^2+z^2}{(y+z)}+ \frac{x^2+z^2}{(x+z)}+ \frac{y^2+x^2}{(y+x)}\\ by QM-AM\ inequality \frac{x^2+y^2}{(x+y)}\geq \frac{x+y}{2},similarly\ for\ others\\ \frac{x^2}{(y+z)}+ \frac{y^2}{(x+z)}+ \frac{z^2}{(y+x)}\geq \frac{y^2+z^2}{(y+z)}+ \frac{x^2+z^2}{(x+z)}+ \frac{y^2+x^2}{(y+x)}\geq \frac{x+y+z}{2}\\ by AM -GM ineq. \\ x+y+z\geq 3\sqrt[3]{xyz}\geq 3\\ so, finally \frac{x^2}{(y+z)}+ \frac{y^2}{(x+z)}+ \frac{z^2}{(y+x)}\geq \frac{3}{2}
i too will prefer the first one :)