Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

If a, b, c be three positive numbers and a+b+c=1 then prove that \[(a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2\ge 33\frac{1}{3}\]

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

@JamesJ Please help me out with this...
Well, by the Cauchy-Schwartz inequality, \[ ( . <1,1,1>)^2 \leq \] \[(1^2 + 1^2 + 1^2) (a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2 \] Play around with that for a while now.
*there should be brackets around all three terms on the right: ( (a+1/a)^2 + (b+1/b)^2 + (c+1/c)^2 )

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

i.e., what the Cauchy-Schwartz inequality says is that if a and b are vectors in \( \mathbb{R}^3 \), then \[ (a \cdot b)^2 \leq ||a||^2||b||^2 \]
I understand what you are saying. I am trying it. I will let you know if I get stuck.
Thanks for the help
Ok. It will come down to you being able to write a smart bound for 1/a + 1/b + 1/c
@JamesJ I used A.M.>H.M. to deal with the part 1/a + 1/b + 1/c. I showed that it is greater than 9, and eventually proved the thing. I hope, this is the right way, is it?
you can't deal with part of function !! 1/a +1/b +1/c can't consider alone.
In minimum solution should consider all term together .
@2bornot2b, yes, sounds right. The Lagrange multiplier method that mahmit has used is also perfectly sound, but we can avoid using such a 'sophisticated' method.

Not the answer you are looking for?

Search for more explanations.

Ask your own question