Positive reals $a,b,c$ satisfy $2a\sqrt{a}+3b\sqrt{b}+18c\sqrt{c}=2022$ and minimize $a^3+b^3+c^3$. Find $a+b+c$.

Question

The positive real numbers $a,b,c$ satisfy $2a \sqrt{a}+3b \sqrt{b}+18c \sqrt{c} = 2022$ and make $a^3+b^3+c^3$ the smallest possible. Find $a+b+c$.

Right now, I'm stuck on where to start. Squaring both sides of the equation didn't seem to get me anywhere. Can I have a direction to head in please? Thanks!

Answer 1

Apply the Cauchy-Schwarz inequality: $2022^2 = (2a\sqrt{a}+ 3b\sqrt{b}+18c\sqrt{c})^2\le (2^2+3^2+18^2)(a^3+b^3+c^3)$. The min value of $a^3+b^3+c^3$ is $\dfrac{2022^2}{2^2+3^2+18^2}=12,132.$ And achieving this value requires that $\dfrac{2}{a\sqrt{a}}=\dfrac{3}{b\sqrt{b}}=\dfrac{18}{c\sqrt{c}}\implies \dfrac{4}{a^3}=\dfrac{9}{b^3}=\dfrac{324}{c^3}= \dfrac{337}{12,132}\implies a =\sqrt[3]{\dfrac{4\cdot 12,132}{337}}=\sqrt[3]{144}$. Can you find the other values? and the sum $a+b+c$?

Answer 2

$ a^3 + b^3 +c^3 \geq 3abc (AM \geq GM) $ with the equality holding true only when $ a^3 = b^3 = c^3$. Since the minimum value of AM($ a^3 + b^3 +c^3$) is GM, $ a^3 = b^3 = c^3$ for this question. so a=b=c

So $23a\sqrt{a} = 2022$

a=250.013
a+b+c=3a=750.039