# Thread: confusion about inequalities involving (x^2 + y^2) ln(x^2 + y^2)

1. ## confusion about inequalities involving (x^2 + y^2) ln(x^2 + y^2)

my book said:
assume 0< x^2 + 2y^2 < 1 then,
0 > (x^2 + y^2)ln( x^2 + 2y^2) >/= (x^2 + y^2)ln(2(x^2 + y^2)) = (x^2 + y^2)ln(x^2 + y^2) + (x^2 + y^2)ln(2)

i understand the first part since the inside of the natural log is less than 1 then the value of ln would be less than 0. and i understand the end which is just the property involving products within a logarithm. what i don't understand is the 2nd inequality. i don't understand how my book knew that adding an additional x^2 inside the logarithm would make ln( x^2 + 2y^2) less than ln(2x^2 + 2y^2). the only constraint we have is that 0< x^2 + 2y^2 < 1 so we know that x^2 is less than 1 but that is all we really know and that fact is too general. i thought the logarithm function is always increasing so making the value inside the ln bigger would make the logarithm bigger and not smaller wouldn't it? is the book wrong or am i missing something? thanks.

2. If $0 then $\ln(a)<\ln(b)$.

How does that apply?

3. well since x^2 + 2y^2 is less than 2(x^2 + y^2), shouldn't ln(x^2 + 2y^2) < ln(2(x^2+y^2))? but the book has it in the opposite direction for some reason saying ln(x^2 + 2y^2) >/= ln(2(x^2 + y^2))

4. If you're accurately relating what the book said, then I would say that because of Plato's post, you can conclude the book's reasoning is wrong at that point. That is, you claim the book is saying this:

Assuming $0 then

$0 > (x^2 + y^2)\ln( x^2 + 2y^2) \ge (x^2 + y^2)\ln(2(x^2 + y^2)) = (x^2 + y^2)\ln(x^2 + y^2) + (x^2 + y^2)\ln(2),$

right?

You're correct. The inequality

$\ln( x^2 + 2y^2) \ge \ln(2(x^2 + y^2))$

doesn't work, even for $0

Counterexample: choose $x=0.5, y=0.6.$ Then

$0 However,

$-0.01858\approx\ln(x^{2}+2y^{2})\not\ge\ln(2(x^{2}+ y^{2}))\approx 0.1213.$