Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

1 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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.