The Idea of Quasiliearisation.

This is a very intrigueing idea due to Russian problem proposer – Fedor Petrov

 We know by AM-GM that ,

2ab\leq a^{2}+b^{2}

Introduce a parameter and get the following:

2ab\leq ta^{2}+\frac{b^{2}}{t}

Then read the last inequality from the other point: for any positive $a,b$ there exist positive t such that

2ab= ta^{2} + \frac{b^{2}}{t}

Here t =\frac{b}{a}

Or, we may write: 2ab =\min(ta^{2}+\frac{b^{2}}{t})

How may this observation help? Put a =\sum (x_{i}^{2}), b =\sum (y_{i}^{2})

Then for appropriate t we have:

2\sqrt{a}\sqrt{b}= ta+\frac{b}{t}=\sum (tx_{i}^{2}+\frac{y_{i}^{2}}{t})\geq\sum (2x_{i}y_{i})

So, we get the Cauchy-Schwarz Inequality. A lot of other inequalities also may be proved by this idea


Prove that for any four nonnegative reals $a,b,c,d$ the following inequality holds-

(ab)^\frac{1}{3} + (cd)^\frac{1}{3} \le ((a+c+b)(a+c+d))^\frac{1}{3}

Source:Proposed at 239 Lyceum Traditional Olympiad (Author : Fedor Petrov)

We have

3(AB)^{\frac{1}{3}}\leq Ax+By+\frac{1}{xy}

And for any positive $A$ and $B$ there exist appropriate $x$ and $y$ ,for which equality holds

x =\frac{(AB)^{\frac{1}{3}}}{A}, y =\frac{(AB)^{\frac{1}{3}}}{B}

Let A = (a+c+b), B = (a+c+d) in terms of the problem. For some positive $x,y$ we have –

3(AB)^{\frac{1}{3}}= Ax+By+\frac{1}{xy}= (a+c+b)x+(a+c+d)y+\frac{1}{xy}=

(a+c+b)x+(a+c+d)y+\frac{1}{x(x+y)}+\frac{1}{y(x+y)}= \big(a(x+y)+bx+\frac{1}{x(x+y)}\big)+\big(c(x+y)+dy+\frac{1}{y(x+y)}\big)\geq 3(ab)^{\frac{1}{3}}+3(cd)^{\frac{1}{3}}


and we are done!


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