Multivariable Inequalitys - Dual Inequality
As a method from《不等式的秘密》(Secrets of Inequalities)
Introduction
In Lagrange multiplier method, its often given that
\[ g(\texttt{v}) = 0\]
and the function
\[ f(\texttt{v}) \geq 0\]
is to be proved.
Many times, g is simpler than f. Then can we swap f and g, in other words, prove that:
\[ f(\texttt{v}) = 0\]
and
\[ g(\texttt{v}) \geq 0\]
?
The answer is yes!
The Idea
WLOG, the following inequalities will all have three variables. However, this method can be easily extended to more variable.
Consider
\[x+y+z=3\]
Show that \[ x + y + z \geq xy + xz + yz \]
This may be hard to prove (is it?) But can we swap what is to be proved and what is given? Well, then we will have
\[ x+ y+ z = xy + xz + yz\]
And we have to show that
\[ x+y+z \geq 3\]
The two inequalities are equivlent. This method is called the dual inequality method. When the original inequality is complex, its dual is often easy.
The Method
If \[A: g(x,y,z) = 0, f(x,y,z) \geq 0 \]
\[ B: g(x,y,z) \geq 0, f(x,y,z) = 0 \]
First, try to show that (which is not hard, use the adjustment method i.e replace x,y,z with tx,ty,tz) \[ A \iff B\]
Second prove \[B\].
Example 1
\[ xyz = 1 \]
\[\frac{1}{2+x} + \frac{1}{2+y} + \frac{1}{2+z} \leq 1\]
Dual:
\[\frac{1}{2+x} + \frac{1}{2+y} + \frac{1}{2+z} = 1\]
\[ xyz \leq 1 \]
Prove dual using substitution:
\[ a = \frac{1}{2+x},b = \frac{1}{2+y},c = \frac{1}{2+z}\]
Then want to show:
\[ (1-2a)(1-2b)(1-2c) \leq abc \]
where
\[ a+b+c = 1 \]
Then want to show:
\[ (a+b-c)(a+c-b)(b+c-a) \leq abc \]
This is schur's inequlity.
Example 2
\[a,b,c>0, ab+bc+ac = 3\]
\[ \sqrt{a+3} + \sqrt{b+3} + \sqrt{c+3} \geq 6\]
Dual:
\[a,b,c>0, ab+bc+ac \leq 3\]
\[ \sqrt{a+3} + \sqrt{b+3} + \sqrt{c+3} = 6\]
Subtitution:
\[ x = \sqrt{a+3}, y = \sqrt{b+3}, z = \sqrt{c+3}\]
Equivlent to:
\[ x+ y+ z = 6\]
\[(x^2-3)(y^2-3)+ (x^2-3)(z^2-3)+ (y^2-3)(z^2-3) \leq 3\]
Simplify:
\[x^2y^2 +y^2z^2 + z^2x^2 + 24 \leq 6(x^2 + y^2 + z^2)\]
Want to make this homogeneous, so:
\[6(x^2y^2 +y^2z^2 + z^2x^2 + 24) \leq (x+y+z)^2(x^2 + y^2 + z^2)\]
\[6(54x^2y^2 +54y^2z^2 + 54z^2x^2 + (x+y+z)^4) \leq 54(x+y+z)^2(x^2 + y^2 + z^2)\]
\[ 324\sum{x^2y^2}+6(\sum{x})^4 \leq 54(\sum x)^2(\sum x^2)^2\]
Now use the total derivitive method.