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.

  • 1
    $\begingroup$ Hint: Let $ x = \log_a b$, etc. Show that this is equivalent to $ xyz = 1, \sum x^2/(2x+1) \geq 1$. So logarithms are no longer involved, and you can try to solve this inequality with others means. $\endgroup$
    – Calvin Lin
    Commented Dec 29, 2023 at 20:30
  • $\begingroup$ "since this is a symmetric inequality, shouldn't the minimum value of the expression be, when a=b=c?" is a common fallacy. $\quad$ Yes the inequality is true, and there are standard ways of proving it, but I don't know what you know or what you've tried. As I stated, please show your work/attempts else MSE will close your question for lack of context. $\endgroup$
    – Calvin Lin
    Commented Dec 30, 2023 at 20:27
  • $\begingroup$ @fikooo: For various reasons it is better to use a single LaTeX formula $ .... $ for an expression and not to split it like $...$ + $...$ + $...$. I have tried to fix that for you (and also use $$ ... $$ for a larger display where appropriate). Please check that I did not introduce any errors. – Btw, I would always use \frac{NUM}{DEN} instead of NUM \over DEN because the scoping is clearer. $\endgroup$
    – Martin R
    Commented Jan 1 at 11:12
  • $\begingroup$ @MartinR everything is fine with your edited version, and thank you for pointing me out these problems! $\endgroup$
    – fikooo
    Commented Jan 1 at 16:13

2 Answers 2


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): $$ \frac{y^2}{2xy+x^2} + \frac{z^2}{2zy+y^2} + \frac{x^2}{2xz + z^2} \ge \frac{(y+z+x)^2}{2xy+x^2 + 2zy+y^2 + 2xz + z^2} = \frac{(x+y+z)^2}{(x+y+z)^2} = 1 \, . $$

  • $\begingroup$ So that was the thing I would have needed to notice...I came across the Cauchy-Schwarz inequality before, but I wouldn't think that it would be the key to this problem, or at least one of its consequences. It seems I've started the new year by learning something new. Thank you for pointing me that out, and Happy New Year! $\endgroup$
    – fikooo
    Commented Jan 1 at 16:18
  • $\begingroup$ @fikooo Can you approach $ \sum x^2 / (2x+1) \geq 1$ via Titu's and see how far you get? You will likely need to show that $ x+y+z \geq 3$. $\endgroup$
    – Calvin Lin
    Commented Jan 1 at 16:29
  • $\begingroup$ @MartinR I'm new to this website, so I'm still getting use to how things work here. I have no problem understanding the solution you have provided, and I accepted it. The only reason I haven't done it earlier is because I didn't know this is how things work here. $\endgroup$
    – fikooo
    Commented Jan 2 at 9:01
  • $\begingroup$ @CalvinLin I showed it, after some struggle, but I've managed to do it. Do I need to show it to you, if yes, where? Writing the solution here would be a little bit messy. $\endgroup$
    – fikooo
    Commented Jan 2 at 10:17
  • $\begingroup$ @fikooo You don't have to. $\quad$ You can post it as a solution. $\endgroup$
    – Calvin Lin
    Commented Jan 2 at 10:58

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

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



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:


which completes our proof, since:

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


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