For example$:$
For $x,y,z>0$ and $xy+yz+zx\geqq 3.$ Prove that$:$
$$\frac{x}{\sqrt{4x + 5y}} + \frac{y}{\sqrt{4y + 5z}} + \frac{z}{\sqrt{4z + 5x}} \geqq 1$$
It is easy and you can see my two proof here.
The main problem of this topic is as follow:
For inequality has form$:$ $$\frac{X}{\sqrt{A}}+\frac{Y}{\sqrt{B}}+\frac{Z}{\sqrt{C}} \geqq k$$
Using $\lceil$ HOLDER!inequality $\rfloor:$
$$\text{LHS} =\sqrt{\frac{(\frac{X}{\sqrt{A}}+\frac{Y}{\sqrt{B}}+\frac{Z}{\sqrt{C}} )^2 (AXW_1^3 +BYW_2^3 +CZW_3^3)}{AXW_1^3 +BYW_2^3 +CZW_3^3}} \geqq \sqrt{\frac{(W_1 X +W_2 Y+W_3 Z)^3}{AXW_1^3 +BYW_2^3 +CZW_3^3}}$$
Our problem become $$(W_1 X +W_2 Y+W_3 Z)^3 \geqq k^2 (AXW_1^3 +BYW_2^3 +CZW_3^3) \,\, -------(1)$$
But to choose $W_1,\,W_2,\,W_3$ is not ussualy easy as the first example. I have no idea to find it$,$ what is your idea$?$
Mr Liu Qian Bao use agl2012 program to find out it!
In the opposite case$,$ $$\frac{X}{\sqrt{A}}+\frac{Y}{\sqrt{B}}+\frac{Z}{\sqrt{C}} \leqq k$$
Using $\lceil$ Cauchy-Schwarz!inequality $\rfloor:$ $$(W_1 +W_2 +W_3) (\frac{X^2}{W_1 A}+\frac{Y^2}{W_2 B} +\frac{Z^2}{W_3 C}) \geqq (\frac{X}{\sqrt{A}}+\frac{Y}{\sqrt{B}}+\frac{Z}{\sqrt{C}})^2$$
Neeed to prove$:$ $$k^2 \geqq (W_1 +W_2 +W_3) (\frac{X^2}{W_1 A}+\frac{Y^2}{W_2 B} +\frac{Z^2}{W_3 C})$$
We have the same problem is find $W_1,\,W_2,\,W_3$$?$
Thank you very much!