hello, it is $ 1\le\frac{a^2+b^2+c^2}{ab+bc+ca}<2$ because $ a^2+b^2+c^2<2ab+2bc+2ca$ is equivalent with $ 0<a(b+c-a)+b(a+c-b)+c(b+a-c)$ and the inequality $ a^2+b^2+c^2 \geq ab+bc+ca$ holds for all real numbers $ a,b,c$, especially when $ a,b,c$ are sides of a triangle. The equal sign holds for $ a=b=c=1$. Sonnhard.

Dr Sonnhard Graubner wrote: $ 0 < a(b + c - a) + b(a + c - b) + c(b + a - c)$ I don't think this is sharp for $ a,b,c$ sides of a triangle.

But I think, that $ b<a+c$ $ a<b+c$ $ c<a+b$ and $ a,b,c>0$ so it's obviously sharp.

But all three of those inequalities cannot be sharp at the same time. What I mean is as $ b\to a+c$, one of the other inequalities becomes less sharp.

$ a^2+b^2+c^2 <k(ab+ac+bc)$ $ (x+y)^2+(y+z)^2+(z+x)^2<k((x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y))=k(x^2+y^2+z^2+3(xy+xz+yz))$ $ 2x^2+2y^2+2z^2+2(xy+xz+yz)<k(x^2+y^2+z^2)+3k(xy+xz+yz)$ Now, k=2 gives a triviality, if k<2 then moving everything over to the L.H.S. gives a positive quadratic in x, which is unbounded positively.

I have'nt checked your solution,rofler :-/ now I have another solution for our problem .Hope it will be correct First of all we have basic ỉnequality :$ ab + bc + ca \leq a^2 + b^2 + c^2 < 2ab + 2bc + 2ca$. This means that $ \frac {a^2 + b^2 + c^2}{ab + bc + ca}$ must be on interval $ [1;2)$ Consider a triangle with three sides of which are $ n,n$ and 1 and a function $ f(n) = \frac {n^2 + n^2 + 1}{n + n + n^2}$ We easily see that f(n) is a continous function and such that $ f(1) = 1$ and $ lim_{n - > + \infty} f(n) = 2$ Therefore $ f(n)$ get all values between $ f(1)$ and $ f( + \infty)$ So $ \frac {a^2 + b^2 + c^2}{ab + bc + ca}$ get all values in interval $ [1;2)$

This si the solution.

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