For $a,b,c \in R$ and $a,b,c>0$ satisfy $a^2+b^2+c^2=27$, minimize $$A=a^3+b^3+c^3$$
For $a,b,c \in R$ and $a,b,c>0$. Minimize $A=a^3+b^3+c^3$
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$\begingroup$
maxima-minima
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2you should perhaps put it in a context. Like what are the prerequisites? Do you know e.g. about Lagrange multipliers? – 2017-02-25
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0@ H. H. Rugh: Ok. And yes i do – 2017-02-25
1 Answers
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By Power-Mean Inequality,
$\left(\dfrac{a^3+b^3+c^3}{3}\right)^{1/3}\ge \left(\dfrac{a^2+b^2+c^2}{3}\right)^{1/2}=3$
$a^3+b^3+c^3 \ge 81$
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0you can use other way ? – 2017-02-25
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0@WordShallow Why do you want to use another way? This is (very) elegant, simple and efficient. – 2017-02-25
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0Do you need an alternate approach ? The other approaches would be longer – 2017-02-25
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2Do you know Jensen's Inequality ? Use it for $f(x)=x^{3/2}$ – 2017-02-25
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0yes, i do -5 more to go... – 2017-02-25
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0Also with a,b,c=3 it becomes 81 so the bound is reached so 81 is the minima. – 2017-03-01