QJEEM16Ma00a6: JEE Main 2016 Mathematics - Class 11 JEEM16Ma00a6 Solution: Let x, y, z be positive real numbers such that x + y + z = 12…

As we know

AM   $\ge$   GM

$\Rightarrow$     ${{3\left( {{x \over 3}} \right) + 4\left( {{y \over 4}} \right) + 5\left( {{z \over 5}} \right)} \over {12}}$    $\ge$    ${\left[ {{{\left( {{x \over 3}} \right)}^3}{{\left( {{y \over 4}} \right)}^4}{{\left( {{z \over 5}} \right)}^5}} \right]^{{1 \over {12}}}}$

$\Rightarrow$    1   $\ge$    ${{{x^3}{y^4}{z^5}} \over {{3^3}{4^4}{5^5}}}$

$\Rightarrow$    x³ y⁴ z⁵   $\le$   3³ . 4⁴ . 5⁵

$\Rightarrow$    x³ y⁴ z⁵   $\le$   (0.1)(600)³

but given that,

x³ y⁴ z⁵ = (0.1) (600)³

$\therefore$    AM   $=$   GM

$\Rightarrow$    All the number are equal.

$\therefore$     ${x \over 3} = {y \over 4} = {z \over 5} = k$

$\Rightarrow$   x $=$ 3k, y = 4k, z = 5k

given that,

        x + y + z $=$ 12

$\Rightarrow$    3k + 4k + 5k $=$ 12

$\Rightarrow$    12k $=$ 12

$\Rightarrow$    k = 1

$\therefore$    x  $=$    3,  y $=$ 4,   z $=$ 5

So, x³ + y³ + z³

$=$ 3³ + 4³ + 5³

$=$ 216