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.

Solution:\color{#D61F06}{\textbf{Solution:}} Write the inequality as cyc(3a+2b2a+b+16a16b)0\sum_{cyc} \left(\frac{3}{a+2b}-\frac{2}{a+b}+\frac{1}{6a}-\frac{1}{6b}\right)\geqslant 0 or cyc(ab)2(2ab)6ab(a+2b)(a+b)0.\sum_{cyc} \frac{(a-b)^2(2a-b)}{6ab(a+2b)(a+b)}\geqslant 0. Since 2ab2122=0,2a-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 16a\frac{1}{6a} and subtracting 16b\frac{1}{6b} helps? Well, let's go back to our initial inequality, 3a+2b+3b+2c+3c+2a2a+b+2b+c+2c+a\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, aa and bb, (since the sum is cyclic) cyc(3a+2b2a+b)0\sum_{cyc} \left(\frac{3}{a+2b}-\frac{2}{a+b}\right)\geqslant 0 or cycba(5a+b)(4a+b)0.\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 kakb\frac{k}{a}-\frac{k}{b} cyclically to the sum (which is just 00) and then if we can find a suitable kk, then we're done right?

So, we start by writing down, cyc(3a+2b2a+b+kakb)0\sum_{cyc} \left(\frac{3}{a+2b}-\frac{2}{a+b}+\frac{k}{a}-\frac{k}{b}\right)\geqslant 0 or cyc(ab)(abk(a+2b)(a+b))ab(a+2b)(a+b)0.\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 (ab)(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 (ab)(a-b), that may help us! Moreover, since abk(a+2b)(a+b)ab-k(a+2b)(a+b) is a polynomial in a,ba,b, if we want to have a factor of (ab)(a-b), we can treat it as a polynomial in aa and it is zero when a=ba=b. Setting a=ba=b, gives a2(16k)=0a^2(1-6k)=0, since a0a\ne 0, we can choose k=16k=\frac{1}{6}.

When k=16k=\frac{1}{6}, the inequality becomes cyc(ab)2(2ab)6ab(a+2b)(a+b)0.\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[13,3]a,b,c \in \left[\frac{1}{\sqrt 3}, \sqrt 3\right], then 52a+3b+52b+3c+52c+3a3a+2b+3b+2c+3c+2a.\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[125,25]a,b,c \in \left[\frac{1}{2\sqrt 5}, 2\sqrt 5\right], then 65a+b+65b+c+65c+a54a+b+54b+c+54c+a.\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[115,152]a,b,c \in \left[\frac{1}{\sqrt {15}}, \frac{\sqrt {15}}{2}\right], then 75a+2b+75b+2c+75c+2a43a+b+43b+c+43c+a.\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}.

Note by ChengYiin Ong
4 weeks, 1 day ago

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