## Monday, April 4, 2016

### Analysis Quiz 22: Proving An Inequality

Let a,b and c be positive real numbers with $abc = 1$, prove that

[MATH]\frac{a}{2+bc}+\frac{b}{2+ca}+\frac{c}{2+ab}\ge  1[/MATH]

Question 1:

Would you see turning the RHS of the inequality of 1 as $abc$ help?

A. Yes.

B. No.

Yes, we could do that, but that wouldn't help, at least it wouldn't give us any hindsight on how to plan strategy or how to proceed.

Question 2:

If you're ask to apply the extended Cauchy-Schwarz inequality to the LHS of the intended inequality, what would be the first thing you would do to the LHS of the intended inequality?

A. Nothing, as we can directly apply the extended Cauchy-Schwarz inequality to the LHS of the intended inequality without a problem.

B. Multiply top and bottom of the fractions by $abc$.

C. Multiply the top and bottom of the fractions by whatever listed in the numerator of that particular fraction.

The extended Cauchy-Schwarz inequality tells us for all positive real x, y, p and q, we have the following inequality that always hold:

[MATH]\frac{x^2}{p}+\frac{y^2}{q}\ge \frac{(x+y)^2}{p+q}[/MATH]

If we want to use that extended Cauchy-Schwarz inequality on [MATH]\frac{a}{2+bc}+\frac{b}{2+ca}+\frac{c}{2+ab}[/MATH], it would be so much nicer if we have the form:

[MATH]\frac{a^2}{2a+abc}+\frac{b^2}{2b+abc}+\frac{c^2}{2c+abc}[/MATH]

and this can be arrived if we do what has been suggested in C.

Question 3:

With the same condition as the original problem, let's say now you're ask to prove [MATH]\frac{(a+b+c)^2}{2(a+b+c)+3abc}\ge 1[/MATH], what strategy would you plan?

A. Expand the numerator.

B. Replace $(a+b+c)^2$ by [MATH]3(ab+bc+ca)[/MATH] since $(a+b+c)^2\ge 3(ab+bc+ca)$.

C. Algebraically manipulate the given equality, which says $abc=1$ by letting [MATH]a=\frac{x}{y},\,b=\frac{y}{z},\,c=\frac{z}{x}[/MATH].

Options A and B wouldn't work, as they only allowed us to walk one step forward, and then put us in great danger as there would be nothing that we could do after we have walked that one step!

We should always think of the substitution skill when we are given $abc=1$ by letting [MATH]a=\frac{x}{y},\,b=\frac{y}{z},\,c=\frac{z}{x}[/MATH]:

[MATH]\frac{a^2}{2a+abc}+\frac{b^2}{2b+abc}+\frac{c^2}{2c+abc}[/MATH]

[MATH]=\frac{a^2}{2a+1}+\frac{b^2}{2b+1}+\frac{c^2}{2c+1}[/MATH]

[MATH]=\frac{\left(\frac{x}{y}\right)^2}{2\left(\frac{x}{y}\right)+1}+\frac{\left(\frac{y}{z}\right)^2}{2\left(\frac{y}{z}\right)+1}+\frac{\left(\frac{z}{x}\right)^2}{2\left(\frac{z}{x}\right)+1}[/MATH]

[MATH]=\left(\frac{x^2}{y^2}\times \frac{y}{2x+y}\right)+\left(\frac{y^2}{z^2}\times \frac{z}{2y+z}\right)+\left(\frac{z^2}{x^2}\times \frac{x}{2z+x}\right)[/MATH]

[MATH]=\frac{x^2}{y(2x+y)}+\frac{y^2}{z(2y+z)}+\frac{z^2}{x(2z+x)}[/MATH]

[MATH]\ge \frac{(x+y+z)^2}{y(2x+y)+z(2y+z)+x(2z+x)}[/MATH]

[MATH]= \frac{(x+y+z)^2}{x^2+y^2+z^2+2(xy+yz+zx)}[/MATH]

[MATH]= \frac{(x+y+z)^2}{(x+y+z)^2}[/MATH]

[MATH]=1[/MATH] (Q.E.D.)

Question 4:

If you are encourage to prove this problem with the Hölder's inequality, do you think you can manage that and successfully prove the problem with just a few steps?

A. Yes.

B. No.