If b + c = ax, c + a = by, a + b = cz, then the value 1/9≤ft[1/(x+1)+…

If $b + c = ax,

c + a = by,

a + b = cz$,

then the value $\frac{1}{9}\left[\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right]$

Put the values of a = 1 , b = 1 and c = 1

Then by putting these values in the given equations the we will get the values of a = 2, b = 2 and c = 2 Now put these values in desired equation,

$\frac{1}{9}\left[\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right]$ = $\frac{1}{9}\left[\frac{1}{2+1}+\frac{1}{2+1}+\frac{1}{2+1}\right]$

= $\frac{1}{9}\left[\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right]$ = \(\frac{1}{9}\) × \(\frac{3}{3}\) = \(\frac{1}{9}\)