# 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

Problem

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 –

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

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

By AM-GM

and we are done!