if a^2+b^2+c^2=3/2 solve 1/a+1/b+1/c<1/(a*b*c)

[alexandrahirsch]

if a^2+b^2+c^2=3/2 solve 1/a+1/b+1/c<1/(a*b*c)

 

[Deleted member 4993]

if a^2+b^2+c^2=3/2 solve 1/a+1/b+1/c<1/(a*b*c)

I do not understand the meaning of the word "solve" in this context.

Did you mean "show that"?

also - if you expand

(a + b + c)²

What do you get?

 

[lookagain]

if a^2+b^2+c^2=3/2 solve 1/a+1/b+1/c<1/(a*b*c)

Even if the problem were changed to the equivalent of:

"Let a, b, c \(\displaystyle \ \ \) belong to the set of positive real numbers.

If a^2 + b^2 + c^2 = 3/2, show/prove 1/a + 1/b + 1/c < 1/(a*b*c),"

it would be false.


It would be false no matter which direction the inequality sign faces.


For example, let \(\displaystyle \ a^2 \ = \ b^2 \ = \ c^2 \ = \ 1/2 \ \ \) so the given equation is satisfied.

Because of the particular set, then \(\displaystyle \ a \ = \ b \ = \ c \ = \ \dfrac{\sqrt{2}}{2}.\)

\(\displaystyle \dfrac{1}{a} \ + \ \dfrac{1}{b} \ + \ \dfrac{1}{c} \ \ \ vs. \ \ \dfrac{1}{abc} \ \ \ \ becomes\)


\(\displaystyle 3\sqrt{2} \ > \ 2\sqrt{2}\)


For a contrasting example, let \(\displaystyle \ a^2 \ = \ \dfrac{11}{8}, \ \ b^2 \ = \ c^2 \ = \ \dfrac{1}{16} \ \ \) so the given equation is satisfied.

Then \(\displaystyle \ a \ = \ \sqrt{\dfrac{11}{8}}, \ \ b \ = \ c \ = \ \dfrac{1}{4}.\)

\(\displaystyle \dfrac{1}{a} \ + \ \dfrac{1}{b} \ + \ \dfrac{1}{c} \ \ \ vs. \ \ \dfrac{1}{abc} \ \ \ \ becomes\)


\(\displaystyle 8.85... \ < \ 13.64...\)