logarithmic inequality with three variables

Question

There's a periodical in my country which comes out every month. It contains math problems gathered from various teachers. Usually the main goal of these problems is to notice something. I've been solving these problems for a year or so, but I came across one which I couldn't figure out. I'm doing it simply because these strategies might become handy one day.

We are given three real numbers a,b,c such that all of them are greater than 1. How to show that the inequality

$${\log_a b\over 2+\log_b a}+{\log_b c\over 2+\log_c b}+{\log_c a\over 2+\log_a c}\ge 1$$

is true?

I've tried to tackle this problem using various methods like rewriting every logarithm so that they can have similar bases. The first thing I've tried was to made the logarithms have the same base in every fraction, so the whole expression become:

$${(\log_a b)^2\over 2\log_a b +1}+{(\log_b c)^2\over 2\log_b c +1}+{(\log_c a)^2\over 2\log_c a +1}\ge 1 \, .$$

After setting $\log_a b=x$, $\log_b c=y$ and $\log_c a=z$, the expression would become

$${x^2\over2x+1}+{y^2\over2y+1}+{z^2\over2z+1}\ge 1 \, .$$

However I don't think this is the right way to tackle this problem since in some sort of way $x$, $y$ and $z$ depend on each other, they can't be completely distinct from each other.

So I tried rewriting the logarithms so that they have base $2$, so it became:

$${(\log_2 b)^2\over 2\log_2 a\log_2 b+(\log_2 a)^2}+{(\log_2 c)^2\over 2\log_2 b\log_2 c+(\log_2 b)^2}+{(\log_2 a)^2\over 2\log_2 a\log_2 c+(\log_2 c)^2}\ge 1 \, .$$

After setting $\log_2 a=x$, $\log_2 b=y$ and $\log_2 c=z$ I got:

$${y^2\over2xy+x^2}+{z^2\over2zy+y^2}+{x^2\over2xz+z^2}\ge 1 \, .$$

This looks a little bit more reasonable for me at least, since $x$, $y$ and $z$ no longer depend on each other, yet in both cases when I tried to have a common denominator I got long polynomials and I couldn't have gone any further. The only thing that simplified things a little bit was that in the first case $xyz$ would equal $1$, but even with this information I couldn't figure it out. I'm trying to this day, I've started solving problems from this book a week ago or so, I have solved vast majority of them, but this one beat me. But I really want to solve it, or at least know how to tackle these kind of inequalities since I was never good at them.

Answer 1

The last version of your inequality $$ \frac{y^2}{2xy+x^2} + \frac{z^2}{2zy+y^2} + \frac{x^2}{2xz + z^2}\ge 1 $$ can be proven with Titu's lemma (which is a direct consequence of the Cauchy-Schwarz inequality): $$ \frac{y^2}{2xy+x^2} + \frac{z^2}{2zy+y^2} + \frac{x^2}{2xz + z^2} \ge \frac{(y+z+x)^2}{2xy+x^2 + 2zy+y^2 + 2xz + z^2} = \frac{(x+y+z)^2}{(x+y+z)^2} = 1 \, . $$

Answer 2

It turns out the first method is also acceptable, which was:

$$\frac{x^2}{2x+1}+\frac{y^2}{2y+1}+\frac{z^2}{2z+1} \ge 1$$

Using Titu's lemma, we get:

$$\frac{x^2}{2x+1}+\frac{y^2}{2y+1}+\frac{z^2}{2z+1} \ge \frac{(x+z+y)^2}{2(x+y+z)+3}$$

If we can show that

$$ \frac{(x+z+y)^2}{2(x+y+z)+3} \ge 1$$

we are done.

Let's name $$t=x+z+y$$

so the inequality becomes

$$\frac{t^2}{2t+3}\ge 1$$

or

$$t^2\ge2t+3$$

which means

$$(t+1)(t-3)\ge0 $$

which implies that $t$ has to be greater than 3, because $t$ cannot be negative. Before we check whether this is true or not, let's see what $xyz$ equals to:

$$\log_a b\log_b c\log_c a=\frac{log_2 a}{log_2 b}\frac{log_2 b}{log_2 c}\frac{log_2 c}{log_2 a}=1$$

As the final step, we'll use AM-GM inequality:

$$\frac{x+y+z}{3}\ge\sqrt[3]{xyz}$$

which completes our proof, since:

$$x+y+z \ge 3$$