Question

If $\frac{x}{a}=\frac{y}{b}=\frac{z}{c},$ prove that

$\frac{x^2+a^2}{x+a}+\frac{y^2+b^2}{y+b}+\frac{z^2+c^2}{z+c}=\frac{(x+y+z{)}^2+(a+b+c{)}^2}{x+y+z+a+b+c}.$

Answer 1

Let $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k,$ so that $x=ak,y=bk,z=ck$;

Then $\frac{x^2+a^2}{x+a}=\frac{a^2{k}^2+a^2}{ak+a}=\frac{(k^2+1)a}{k+1};$

$\therefore \frac{x^2+a^2}{x+a}+\frac{y^2+b^2}{y+b}+\frac{z^2+c^2}{z+c}=\frac{(k^2+1)a}{k+1}+\frac{(k^2+1)b}{k+1}+\frac{(k^2+1)c}{k+1}$

$=\frac{(k^2+1)(a+b+c)}{k+1}$

$=\frac{k^2(a+b+c{)}^2+(a+b+c{)}^2}{k(a+b+c)+(a+b+c)}$

$=\frac{(ka+kb+kc{)}^2+(a+b+c{)}^2}{(ka+kb+kc)+a+b+c}$

$=\frac{(x+y+z{)}^2+(a+b+c{)}^2}{x+y+z+a+b+c}.$