Let x, y, z be positive real numbers su | vedclass

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Question

$x+y+z=12$

$AM \ge GM$

$\frac{{3\left( {\frac{x}{3}} \right) + 4\left( {\frac{y}{4}} \right) + 5\left( {\frac{z}{5}} \right)}}{{12}} \ge $

$\

Vedclass · Accepted Answer

$216$

Vedclass · Answer

$x+y+z=12$

$AM \ge GM$

$\frac{{3\left( {\frac{x}{3}} ight) + 4\left( {\frac{y}{4}} ight) + 5\left( {\frac{z}{5}} ight)}}{{12}} \ge $

$\sqrt[{12}]{{{{\left( {\frac{x}{3}} ight)}^3}{{\left( {\frac{y}{4}} ight)}^4}{{\left( {\frac{z}{5}} ight)}^5}}}$

$\frac{{{x^3}{y^4}{z^5}}}{{{3^3}{4^4}{5^5}}} \le 1$

${x^3}{y^4}{z^5} \le {3^3}{.4^4}{.5^5}$

${x^3}{y^4}{z^5} \le \left( {0.1} ight){\left( {600} ight)^3}$

But, given ${x^3}{y^4}{z^5} = \left( {0.1} ight){\left( {600} ight)^3}$

$	herefore $ all the number are equal

$	herefore \frac{x}{3} = \frac{y}{4} = \frac{z}{5}\left( { = k} ight)$

$x = 3k;y = 4k;z = 5k$

$x + y + z = 12$

$3k + 4k + k = 12$

$k = 1$

$	herefore x = 3;y = 4;z = 5$

$	herefore {x^3} + {y^3} + {z^3} = 216$

Let $x, y, z$ be positive real numbers such that $x + y + z = 12$ and $x^3y^4z^5 = (0. 1 ) (600)^3$. Then $x^3 + y^3 + z^3$ is equal to

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