How prove this inequality with $\sqrt{\sum_{cyc}\frac{a^2}{b^2}}+\sqrt{3}\ge\sqrt{\sum\frac{a}{b}}+\sqrt{\sum\frac{b}{a}}$

Question

Let $a,b,c>0$, show that $$\sqrt{\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}}+\sqrt{3}\ge\sqrt{\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}}+\sqrt{\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}}$$ I tried C-S, AM-GM, Holder and more, but without success.

following maybe is idea:$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=x\Longrightarrow x^2=\sum_{cyc}\dfrac{a^2}{b^2}+2\sum_{cyc}\dfrac{b}{a}$$ then How can prove this inequality?

Answer 1

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

Hence, $xyz=1$ and we need to prove that $$\sqrt{x^2+y^2+z^2}+\sqrt{3}\geq\sqrt{x+y+z}+\sqrt{xy+xz+yz}.$$ or $$x^2+y^2+z^2+3+2\sqrt{3(x^2+y^2+z^2)}\geq x+y+z+xy+xz+yz+2\sqrt{(xy+xz+yz)}$$ or $$x^2+y^2+z^2+3-2(xy+xz+yz)+2\sqrt{3(x^2+y^2+z^2)}-2(x+y+z)+$$ $$+\left(\sqrt{xy+xz+yz}-\sqrt{x+y+z)}\right)^2\geq0,$$ which is true because by C-S $$\sqrt{3(x^2+y^2+z^2)}=\sqrt{(1^2+1^2+1^2)(x^2+y^2+z^2)}\geq x+y+z$$ and by Schur and by AM-GM we obtain: $$x^2+y^2+z^2+3=\sum_{cyc}\left(\left(x^{\frac{2}{3}}\right)^3+x^{\frac{2}{3}}y^{\frac{2}{3}}z^{\frac{2}{3}}\right)\geq$$ $$\geq\sum_{cyc}\left(x^{\frac{4}{3}}y^{\frac{2}{3}}+x^{\frac{2}{3}}y^{\frac{4}{3}}\right)\geq\sum_{cyc}2xy=2(xy+xz+yz).$$ Done!