# Inequality with mean inequality

If $$a,b,c > 0$$ and $$a+b+c = 18 abc$$, prove that $$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{c^{2}}}\geq 9$$

I started writing the left member as $$\frac{\sqrt{3a^{2}+ 1}}{a}+ \frac{\sqrt{3b^{2}+ 1}}{b} + \frac{\sqrt{3c^{2}+ 1}}{c}$$ and I applied AM-QM inequality, but I obtain something with $$\sqrt[4]{3}$$.

• Use Minkowski's Inequality and finish off with rearrangement inequality, similar to solutions to Problem 682 here: artofproblemsolving.com/community/c6h299899p3084827 Nov 18, 2020 at 13:17
• If I am not wrong, threads containing answers to this question got deleted $2$ times. Nov 18, 2020 at 13:29
• Is there a typo in third radical, which now matches second? [should its $b^2$ really be $c^2$] Nov 18, 2020 at 13:32

$$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}$$ $$= \sqrt{ 9 + \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} + 2\sum_{cyc} {\sqrt{ \left(3 + \frac{1}{a^{2}}\right) \left(3 +\frac{1}{b^{2}}\right)}} }$$ $$\tag{By C-S}\geqslant \sqrt{9 + \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} + 2\sum_{cyc}{3 + \frac{1}{ab}}}$$ $$= \sqrt{9 + \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} + 54}$$ $$\geqslant \sqrt{9 +18 + 54} =\sqrt{81} = 9$$ Which uses $$\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = 18$$ and $$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \geqslant \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$$.

• can you clarify the C-S step? Nov 18, 2020 at 14:13
• nice!(+1) but @cosmo5 that step is right and self explanatory Nov 18, 2020 at 15:06
• @cosmo5 If you are thinking of applying C-S on $\sum_{cyc}{\sqrt{\left(3 +\frac{1}{a^{2}}\right)\left(3 +\frac{1}{b^{2}}\right)}}$, think of applying C-S to each term in that sum, not on entire sum. Nov 18, 2020 at 15:18

let $$x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}$$ $$xy+yz+zx=18 \tag{given}$$ $$\color{red}{x+y+z\ge 3\sqrt{6}} \tag1$$

notice that $${(x-\sqrt{6})}^2\ge 0\Rightarrow \sqrt{3+x^2}\ge \frac{\sqrt{6}}{3}x+1$$

Thus $$\sum \sqrt{3+x^2}\ge \sum \frac{\sqrt{6}}{3}x+1\ge\frac{\sqrt{6}}{3}\color{red}{3\sqrt{6}}+3=9$$

Note the inequality marked $$(1)$$ is left as an exercise

• Very nice! ($+1$) Only the last equality should be inequality. Nov 18, 2020 at 15:13

From Minkowski inequality: $$\sqrt{x^2 + a^2} + \sqrt{y^2 + b^2} \ge \sqrt{(x + y)^2 + (a + b)^2}$$

applied to the first two terms, with the notation of @Albus, we get: $$\sqrt{\sqrt{3}^2 +x^2} + \sqrt{\sqrt{3}^2 +y^2} \ge \sqrt{(2\sqrt{3})^2 + (x+ y)^2}$$. Let's denote RHS of the inequality by $$X$$.

Then, apply Minkowski inequality again to $$X + \sqrt{3 + z^2}$$, to obtain: $$\sqrt{27 + (x + y + z)^2} \ge 9$$, which is equivalent to (1) from @Albus Dumbledore's answer.

Hints:

Put Let $$a=\frac{x}{3\sqrt2},$$ $$b=\frac{y}{3\sqrt2}$$ and $$c=\frac{z}{3\sqrt2}.$$

And $$x=\tan(u)$$,$$y=\tan(v)$$,$$z=\tan(w)$$ with $$u+v+w=\pi$$

Then the function $$f(u)=\sqrt{3+\frac{18}{\tan(u)^2}}$$ is convex on $$(0,\pi)$$ see here

Remains to apply Jensen's inequality .

• How do you get the idea to conver to trigonometry? In a middle / high-school contest, I find taking the second derivative of f(u) a bit challenging. Dec 27, 2020 at 16:36