$a;b;c\in \mathbb{R}^+$. Prove : $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a+b+c}{\sqrt{a^2+b^2+c^2}} \geq 3+\sqrt{3}$

Question

$a;b;c\in \mathbb{R}^+$. Prove : $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a+b+c}{\sqrt{a^2+b^2+c^2}} \geq 3+\sqrt{3}$

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I have proved that : $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3\sqrt[3]{\frac{abc}{bca}}=3$

And : $\frac{a+b+c}{\sqrt{a^{2}+b^{2}+c^{2}}}\leq \frac{\sqrt{3}(a+b+c)}{a+b+c}=\sqrt{3}$

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Answer 1

use this known lema: $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{9(a^2+b^2+c^2)}{(a+b+c)^2}$$ proof:By applying the know inequality $$(x+y+z)^3\ge\dfrac{27}{4}(x^2y+y^2z+z^2x+xyz)$$ for $x=\dfrac{a}{b},y=\dfrac{b}{c},z=\dfrac{c}{a}$,we get $$\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)^3\ge\dfrac{27}{4}\left(\dfrac{a^3+b^3+c^3}{abc}+1\right)$$ it suffices to prove that $$\dfrac{a^3+b^3+c^3}{abc}+1\ge\dfrac{108(a^2+b^2+c^2)^3}{(a+b+c)^6}$$ or $$\dfrac{(a+b+c)(a^2+b^2+c^2-ab-bc-ac)}{abc}+4\ge\dfrac{108(a^2+b^2+c^2)^3}{(a+b+c)^6}$$ Using now the obvious inequality $3abc(a+b+c)\le (ab+bc+ac)^2$,we have

$$\dfrac{a+b+c}{abc}=\dfrac{3(a+b+c)^2}{3abc(a+b+c)}\ge\dfrac{3(a+b+c)^2}{(ab+bc+ac)^2}$$ and hence,it is enough to check that $$\dfrac{3(a+b+c)^2(a^2+b^2+c^2-ab-bc-ac)}{(ab+bc+ac)^2}+4\ge\dfrac{108(a^2+b^2+c^2)^3}{(a+b+c)^6}$$ let $$t=\dfrac{3(a^2+b^2+c^2)}{(a+b+c)^2},1\le t<3$$ then inequality is equivalent to $$\dfrac{54(t-1)}{(3-t)^2}+4\ge 4t^3$$ or $$(t-1)(9-6t-8t^2+10t^3-2t^4)\ge 0$$ this is true because $$9-6t-8t^2+10t^3-2t^4=2(3+3t-t^2)(t-1)^2+3>0$$

This lema prove by done.

so you can let $$x=\dfrac{a^2+b^2+c^2}{(a+b+c)^2}\ge\dfrac{1}{3}$$ so use this lemma, it suffices to prove that $$9x+\dfrac{1}{\sqrt{x}}\ge 3+\sqrt{3}$$ let $f(x)=9x+\dfrac{1}{\sqrt{x}},x\ge\dfrac{1}{3}$ then we have $$f'(x)=9-\dfrac{1}{2x^{\frac{3}{2}}}>0$$ so $$f(x)\ge f(\dfrac{1}{\sqrt{3}})=3+\sqrt{3}$$