Question

If a, b, x, y are positive natural numbers such that $\frac{1}{x}+\frac{1}{y}=1$ and $\frac{a^x}{x}+\frac{b^y}{y}$ is-

• A. $\le ab$
• B. $\ge ab$
• C. $=ab$
• D. Can't be found out

Answer 1

Answer: (B)

$\ge ab$

Consider the opposite numbers
$a^x,a^x,.....$ ky times and $b^y,b^y,.....kx$ times
$AM=\frac{\{a^x+a^x+.....kytimes\}+\{b^y+b^y+.....kxtimes\}}{kx+ky}=\frac{ky{a}^x+kx{b}^y}{k(x+y)}=\frac{y{a}^x+x{b}^y}{(x+y)}$
$GM={\left\{\right(a^x.a^x....kytimes\left)\right(b^y.b^y....kxtimes\left)\right\}}^{\frac{1}{k(x+y)}}$
$(a^{x\left(ky\right)}.b^{y\left(kx\right)}{)}^{\frac{1}{k(x+y)}}=(ab{)}^{\frac{kxy}{k(x+y)}}=(ab{)}^{\frac{xy}{(x+y)}}.....(i)$
As $\frac{1}{x}+\frac{1}{y}=1,\frac{x+y}{xy}=1,i.e,x+y=xy$
$\therefore \left(i\right)becomes\frac{y{z}^x+x{b}^y}{xy}\ge ab$ or $\frac{a^x}{x}+\frac{b^y}{y}\ge ab$
Hence, option 'B' is correct.