ISI (1)

Let $a+b+c=0$ , we need to find the value of $\large P= \frac{a^2}{2a^2+bc} + \frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}=?$

Since $a+b+c=0\implies$

$b= -(a+c) \implies$

$bc=-c(a+c) \implies$

$2a^2+bc=2a^2-c(a+c)$

$=a^2+a^2-ac-c^2=$

$(a^2-c^2)+a(a-c)=$

$(a-c)(a+c)+a(a-c)=$

$(a-c)(2a+c)=(a-c)(a-b).$

$\large\implies\frac{a^2}{2a^2+bc}=\frac{a^2}{(a-b)(a-c)}.$

Similarly, we can show that $\large\frac{b^2}{2b^2+ac}=\frac{b^2}{(b-a)(b-c)} \text { and }$

$\large\frac{c^2}{2c^2+ab}=\frac{c^2}{(c-a)(c-b)}.$

Therefore: $\large P= \frac{a^2}{2a^2+bc} + \frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}= \frac{a^2}{(a-b)(a-c)}+ \frac{b^2}{(b-a)(b-c)} +\frac{c^2}{(c-a)(c-b)} .$

$\large P= \frac{1}{(a-b)}\Big( \frac{a^2}{(a-c)}+\frac{b^2}{(b-c)}\Big)+\frac{c^2}{(c-a)(c-b)}.$

$\large P=\frac{ab-ca-cb}{(c-a)(c-b)} + \frac{c^2}{(c-a)(c-b)}=\frac{ab-ca-cb+c^2}{(c-a)(c-b)}.$

$\implies \large P= \frac{(c-a)(c-b)}{(c-a)(c-b)}= 1$ .