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$a$, $b$ and $c$ are positives such that $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \le 3$. Calculate the maximum value of $$\large \frac{1}{\sqrt{a^2 - ab + 3b^2 + 1}} + \frac{1}{\sqrt{b^2 - bc + 3c^2 + 1}} + \frac{1}{\sqrt{c^2 - ca + 3a^2 + 1}}$$

I want to ask if there are any other solutions that are more practical, please. This was taken directly from an exam I recently took. I have posted a solution below if anyone to check out.

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We can use also the following way.

By C-S and Tangent Line method we obtain: $$\sum_{cyc}\frac{1}{\sqrt{a^2-ab+3b^2+1}}=\sum_{cyc}\frac{2}{\sqrt{(3+1)(a^2-ab+3b^2+1}}\leq$$ $$\leq2\sum_{cyc}\frac{1}{\sqrt{3(a^2-ab+3b^2)}+1}\leq\frac{2}{(3+1)^2}\sum_{cyc}\left(\frac{3^2}{\sqrt{a^2-ab+3b^2}}+\frac{1^2}{1}\right)=$$ $$=\frac{3}{8}+\frac{9}{8}\sum_{cyc}\left(\frac{1}{\sqrt{3(a^2-ab+3b^2)}}-\frac{1}{18a}-\frac{5}{18b}\right)+\frac{9}{8}\sum_{cyc}\left(\frac{1}{18a}+\frac{5}{18b}\right)\leq$$ $$\leq\frac{3}{8}+\frac{3}{8}\sum_{cyc}\frac{1}{a}\leq\frac{3}{8}+\frac{9}{8}=\frac{3}{2}.$$ The equality occurs for $a=b=c=1,$ which says that we got a maximal value.

I got $$\frac{1}{\sqrt{3(a^2-ab+3b^2)}}\leq\frac{1}{18a}+\frac{5}{18b}$$ by the following way.

Let $a=xb$.

Now, we'll choose $k$ and $m$ such that a graph of $$g(x)=\frac{k}{x}+m$$ is a tangent to the graph $$f(x)=\frac{1}{\sqrt{3(x^2-x+3)}}$$ for $x=1$,

for which we need $$f(1)=g(1)$$ and $$f'(1)=g'(1).$$ After solving of this system we obtain $k=\frac{1}{18}$ and $m=\frac{5}{18}.$

Now, easy to show that the inequality $$\frac{1}{\sqrt{3(x^2-x+3)}}\leq\frac{1}{18x}+\frac{5}{18}$$ is true for all $x>0$.

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  • $\begingroup$ What is the Tangent Line Method? I have never heard of that. $\endgroup$ – Lê Thành Đạt Jun 3 '19 at 13:04
  • $\begingroup$ Firstly, you saw it already in my solution. Also, I have no possibility to explain it now. In evening. OK? $\endgroup$ – Michael Rozenberg Jun 3 '19 at 14:01
  • $\begingroup$ @Lê Thành Đạt I added something. See now. $\endgroup$ – Michael Rozenberg Jun 3 '19 at 17:36
  • $\begingroup$ Oh, so it is like Unidentified Variables, right? $\endgroup$ – Lê Thành Đạt Jun 4 '19 at 5:51
  • $\begingroup$ Yes, we can say so. $\endgroup$ – Michael Rozenberg Jun 4 '19 at 5:58
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We have that $$a^2 - ab + 3b^2 + 1 = (a^2 - ab + b^2) + b^2 + (b^2 + 1) \ge ab + b^2 + 2b = b(a + b + 2)$$

$$\implies \frac{1}{\sqrt{a^2 - ab + 3b^2 + 1}} \le \frac{1}{\sqrt{b(a + b + 2)}} = \frac{2}{\sqrt{4b(a + b + 2)}} \le \frac{1}{4b} + \frac{1}{a + b + 2}$$

Using Titu's Lemma for $(1, 1, 2)$ and $(a, b, 2)$, we have that $\dfrac{16}{a + b + 2} \le \dfrac{1}{a} + \dfrac{1}{b} + 2$.

$$\implies \frac{1}{4b} + \frac{1}{a + b + 2} \le \frac{1}{4b} + \frac{1}{16}\left(\frac{1}{a} + \frac{1}{b} + 2\right) = \frac{1}{16}\left(\frac{1}{a} + \frac{5}{b} + 2\right)$$

$$\implies \frac{1}{\sqrt{a^2 - ab + 3b^2 + 1}} \le \frac{1}{16}\left(\frac{1}{a} + \frac{5}{b} + 2\right)$$

In conclusion,

$$\sum_{cyc}\frac{1}{\sqrt{a^2 - ab + 3b^2 + 1}} \le \frac{1}{16}\left(\frac{6}{a} + \frac{6}{b} + \frac{6}{c} + 6\right)$$

$$ = \frac{3}{8}\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 1\right) \le \frac{3}{8}\left(3 + 1\right) = \frac{3}{2}$$

The equality sign happens when $a = b = c = 1$.

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