An Interesting Type of Inequalities

(Vasile Cirtoaje) If \(a,b,c\in \left[\frac{1}{\sqrt 2}, \sqrt 2\right],\) show that \[\frac{3}{a+2b}+\frac{3}{b+2c}+\frac{3}{c+2a}\geqslant \frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\] holds.

$\color{#D61F06}{\textbf{Solution:}}$ Write the inequality as $\sum_{cyc} \left(\frac{3}{a+2b}-\frac{2}{a+b}+\frac{1}{6a}-\frac{1}{6b}\right)\geqslant 0$ or $\sum_{cyc} \frac{(a-b)^2(2a-b)}{6ab(a+2b)(a+b)}\geqslant 0.$ Since $2a-b \geqslant 2 \cdot \frac{1}{\sqrt 2}-\sqrt 2=0,$ the inequality holds obviously. $\quad \blacksquare$

However, you may wonder, how in the world do we know that adding $\frac{1}{6a}$ and subtracting $\frac{1}{6b}$ helps? Well, let's go back to our initial inequality, $\frac{3}{a+2b}+\frac{3}{b+2c}+\frac{3}{c+2a}\geqslant \frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}$ the motivation behind the solution is that if we can somehow turn it to a cyclic sum of non-negative terms, (which contain some squares in them) then we're done! So, intuitively, let's try to just move the RHS terms to the LHS, and focus on just two variables, $a$ and $b$, (since the sum is cyclic) $\sum_{cyc} \left(\frac{3}{a+2b}-\frac{2}{a+b}\right)\geqslant 0$ or $\sum_{cyc} \frac{b-a}{(5a+b)(4a+b)}\geqslant 0.$ No good here, as we cannot guarantee each term of the cyclic sum to be non-negative.

Now, why don't we manipulate the sum a little? Let's add $\frac{k}{a}-\frac{k}{b}$ cyclically to the sum (which is just $0$) and then if we can find a suitable $k$, then we're done right?

So, we start by writing down, $\sum_{cyc} \left(\frac{3}{a+2b}-\frac{2}{a+b}+\frac{k}{a}-\frac{k}{b}\right)\geqslant 0$ or $\sum_{cyc} \frac{(a-b)(ab-k(a+2b)(a+b))}{ab(a+2b)(a+b)}\geqslant 0.$ Here, we can see that there's a $(a-b)$ term which is not always non-negative, but the square of it is non-negative by the trivially inequality. Thus, if we can have an extra factor of $(a-b)$, that may help us! Moreover, since $ab-k(a+2b)(a+b)$ is a polynomial in $a,b$, if we want to have a factor of $(a-b)$, we can treat it as a polynomial in $a$ and it is zero when $a=b$. Setting $a=b$, gives $a^2(1-6k)=0$, since $a\ne 0$, we can choose $k=\frac{1}{6}$.

When $k=\frac{1}{6}$, the inequality becomes $\sum_{cyc} \frac{(a-b)^2(2a-b)}{6ab(a+2b)(a+b)}\geqslant 0.$

Using this method, we can also prove some other similar inequalities:

1. Prove that if $a,b,c \in \left[\frac{1}{\sqrt 3}, \sqrt 3\right]$, then $\frac{5}{2a+3b}+\frac{5}{2b+3c}+\frac{5}{2c+3a}\leqslant \frac{3}{a+2b}+\frac{3}{b+2c}+\frac{3}{c+2a}.$

2. Prove that if $a,b,c \in \left[\frac{1}{2\sqrt 5}, 2\sqrt 5\right]$, then $\frac{6}{5a+b}+\frac{6}{5b+c}+\frac{6}{5c+a}\geqslant \frac{5}{4a+b}+\frac{5}{4b+c}+\frac{5}{4c+a}.$

3. Prove that if $a,b,c \in \left[\frac{1}{\sqrt {15}}, \frac{\sqrt {15}}{2}\right]$, then $\frac{7}{5a+2b}+\frac{7}{5b+2c}+\frac{7}{5c+2a}\leqslant \frac{4}{3a+b}+\frac{4}{3b+c}+\frac{4}{3c+a}.$