a) Replace $\sqrt{y+z}=a, \sqrt{z+x}=b, \sqrt{x+y}=c$. Then because $a^2, b^2, c^2$ are sides of a triangle we obtain that $a, b, c$ are sides of an acute triangle. The inequality becomes $$a+b+c>4R\quad (*)$$where $R$ is the circumradius. But that inequality is right because from the acuteness of the triangle we have $a+b>2R+\sqrt{4R^2-c^2}$ whence $$a+b+c>2R+\sqrt{4R^2-c^2}+c>4R.$$b) We should prove that for any $\varepsilon>0$ there exists an acute triangle with sidelengths $a, b, c$ such that $$a+b+c<(4+\varepsilon)R.$$Draw a chord $AB$ in the circle of radius $R$ so that $AB<\varepsilon R$ and mark the point $C$ on the circle in such a way that the triangle $ABC$ is isosceles and acute. Then obviously $$a+b+c<2R+2R+\varepsilon R=(4+\varepsilon)R.$$

Thank you very much! To both of you. In my alternative solution, I use Karamata for the strictly concave function $f(x)=\sin x$ on the interval $\left[0,\frac{\pi}2\right].$

parmenides51 wrote: Let $k > 2$ be a real number. a) Prove that for all positive real numbers $x,y$ and $z$ the following inequality holds: $$\sqrt{x + y }+\sqrt{y + z }+\sqrt{z + x} > 2\sqrt{\frac{(x + y)(y + z)(z + x)}{xy + yz + zx}}$$b) Prove that there exist positive real numbers $x, y$ and $z$ such that $$\sqrt{x + y }+\sqrt{y + z}+\sqrt{z + x} <k\sqrt{\frac{(x + y)(y + z)(z + x)}{xy + yz + zx}}$$Leonard Giugiuc My alternative solution, which solution couldn't be featured for juniors. Published by the Professor Alexander Bogomolny, on 29 May 2018, here https://www.cut-the-knot.org/m/Algebra/LeoInSquareRoots.shtml

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