Properties of Triangles 2 Question 9
9. Show that for any triangle with sides $a, b, c$ $3(a b+b c+c a) \leq(a+b+c)^{2} \leq 4(a b+b c+c a)$.
$(1979,3 M)$
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Solution:
- By using triangular inequality,
$c<a+b$
$c^{2}<c a+a b$
Similarly, $a^{2}<a b+a c$ and $b^{2}<b c+a b$
$$ \begin{array}{lc} \therefore & a^{2}+b^{2}+c^{2}<2 a b+2 b c+2 c a \\ \Rightarrow & \left(a^{2}+b^{2}+c^{2}\right)+2 a b+2 b c+2 c a<4(a b+b c+c a) \\ \Rightarrow & (a+b+c)^{2}<4(a b+b c+c a) \end{array} $$
Now, using AM-GM inequality in $a, b$ and $c$, we get
$$ \begin{aligned} & \frac{a^{2}+b^{2}}{2} \geq a b, \frac{b^{2}+c^{2}}{2} \geq b c \text { and } \frac{c^{2}+a^{2}}{2} \geq c a \\ \Rightarrow & a^{2}+b^{2}+c^{2} \geq a b+b c+c a \\ \Rightarrow & a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a \geq 3(a b+b c+c a) \\ \Rightarrow & (a+b+c)^{2} \geq 3(a b+b c+c a) \end{aligned} $$
From Eqs. (i) and (ii), we get
$$ 3(a b+b c+c a) \leq(a+b+c)^{2}<4(a b+b c+c a) $$
