Which of the following statements are true?
(a) if $a,b$ and $c$ are the sides of a triangle , then $(ab+bc+ca)/(a^2+b^2+c^2)≥(1/2)$
(b) if $a,b$ and $c$ are the sides of a triangle , then $(ab+bc+ca)/(a^2+b^2+c^2)≤ 1$
(c) both statements above are true for all triples $(a,b,c)$ of strictly positive real numbers.
How can I do this?
My thoughts:
(b) true , (a) true, (c) false
Best Answer
$$(a-b)^2+(b-c)^2+(c-a)^2\ge 0$$ $$\implies a^2+b^2+c^2\ge ab+bc+ca\iff \frac{ab+bc+ca}{a^2+b^2+c^2}\le1$$
If $c\ge a-b, c^2\ge (a-b)^2=a^2+b^2-2ab$ which is true for any triangle
Similarly, $a^2\ge b^2+c^2-2bc$
and $b^2\ge c^2+a^2-2ca$
Adding we get, $$a^2+b^2+c^2\ge 2(a^2+b^2+c^2)-2(ab+bc+ca)$$
$$\iff 2(ab+bc+ca)\ge a^2+b^2+c^2\iff \frac{ab+bc+ca}{a^2+b^2+c^2}\ge \frac12$$
So, $(a),(b)$ are true, not $(c)$ as we need the difference of any two of $a,b,c$ to be $\ge$ the other.