Question

Prove that maximum value of a^2b^3c^4 subject to a+b+c=18 is 4^26^38^4

Answer 1

a+b+c = 18

Let me rewrite this statement as:

\[\LARGE \frac{a}{2}+\frac{a}{2}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4} = 18\]

Apply AM-GM inequality to the above expression to get:
\[\frac{18}{9} \ge (\frac{a^2b^3c^4}{2^23^34^4})^{1/9}\]
\[\large \implies a^2b^3c^4 \le 2^9*2^2*3^3*4^4 = 2^{9+2+8}3^{3}=2^{19}3^3\]

Your question asks us to prove \(\large 4^{2}6^{3}8^{4} = 2^{4}*2^33^3*2^{12}=2^{19}*3^{3}\) which is what is proved.

Answer 2

that was some geniusness ! B)