An inequality question?

Given that a1, a2, b1, b2, m, M are positive numbers satisfying m≦(bk/ak)≦M (k = 1,2).

(a) Show that (bk^2) + mM(ak^2) ≦ (m+M)akbk for k = 1, 2.
(b) Hence, show that [(a1^2)+(a2^2)][(b1^2)+(b2^2)] / [(a1b1+a2b2)^2] ≦ [(m+M)^2]/4mM

Remark: ak is a number, it does not mean a times k but it just behaves like a member in a sequence. Similar notation for bk, b1, b2, etc.

Thank you!

Answers:
Oooooooh, I always wondered what chemists did for fun !!

Give Charles the points,he's taking you seriously...

M :D
Well, here's part a. Obviously, I solved this from the bottom up -- it's easier to follow that way, I think.

Let bk/ak = R (subscript suppressed, since the case for both is identical). Then we have
0 < m ≤ R ≤ M.

Clearly, 1 ≤ M/R and so
(R-m) ≤ (R-m)M/R
R ≤ (R-m)M/R + m
R² ≤ (R-m)M + mR
R² + mM ≤ RM + mR
R² + mM ≤ R(M+m)
(bk/ak)² + mM ≤ (bk/ak)(M+m)
(bk)² + mM(ak)² ≤ bkak(M+m), as desired.
.Your world seems remarkably different to mine... And inherently complex!
Jesus Christ go out and get a life. Does anybody give a f.u.c.k. what the answer is, will the world stop turning without an answer, thats it I'm of, going to the pub, goodbye.
I feel as if I've fallen down Lewis Carrol's rabbit hole
What a change from a 'womens' lib question'!

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