Question

If a, b, c are the sides of a triangle then which of the following hold good?

• A. $a^2+b^2+c^2>ab+bc+ca$
• B. $\frac{a^2+b^2+c^2}{ab+bc+ca}$ lies between 1 and 2
• C. $a^3+b^3+c^3>3abc$
• D. $3(ab+bc+ca)\le (a+b+c{)}^2\le 4(bc+ca+ab)$

Answer 1

Answer: (A), (B), (C), (D)

$3(ab+bc+ca)\le (a+b+c{)}^2\le 4(bc+ca+ab)$

(a), (b), (c), (d)
(a) We have proved in rule two by two, that
$a^2+b^2+c^2>ab+bc+ca\dots \dots \left(1\right)$

(b) From above it follows that the given fraction is greater than 1.
Again we know that in a triangle by cosine formula
$b^2+c^2-a^2=2bccosA<2bc\because cosA<1$
$c^2+a^2-b^2=2cacosB<2ca$
$a^2+b^2-c^2=2abcosC<2ab$ Adding
$a^2+b^2+c^2<2(ab+bc+ca)\dots \dots \left(A\right)$
$\therefore \frac{a^2+b^2+c^2}{ab+bc+ca}<2$
Hence the given fraction lies between 1 and 2.

(c) Again $a^3+b^3+c^3-3abc$
$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$=(+)(+)i.e.>0$ by (1)
$a^3+b^3+c^3>3abc$
Note: It can be proved by A.M. \geq G.M.

(d) Adding $2\sum ab$ to both sides of (1), we get
$(a+b+c{)}^2>3(ab+bc+ca)\dots \dots (B)$
or $3\sum ab\le 4(ab+bc+ca)\dots \dots \left(C\right)$
Hence we prove result (d) from (B) and (C).