practice problems

Q1. Prove that, for a, b, c ∈ ℝ⁺, a/b + b/c + c/a ≥ (a + b + c)² / (ab + bc + ca).
S1.
We will use Titu's lemma to prove it.
First let's prove Titu's using C-S.

C-S says for any real numbers:
(a1b1 + a2b2 ... anbn)^2 <= (a1^2+ a2^2 .. an^2).(b1^2+b2^2...bn^2)
Let's use p1 = a1/sqrt(b1)... with addl. constraint that b1,b2.. are positive reals.
q1 = sqrt(b1)...
It becomes:
(p1 + p2 .. pn)^2 <= (p1^2/q1 ..)(q1+q2...qn)
=>
(p1+p2..pn)^2/(q1+q2..qn) <= p1^2/q1 + p2^2/q2...
This is Titu's lemma.

Now, for the given question.
Make LHS like this:
a^2/ab + b^2/bc + c^2/ac

In Titu's lemma, let's use:
p1 = a, p2 = b, p3 = c
q1 = ab, q2 = bc, q3 = ac

So it becomes:
(a+b+c)^2/(ab+bc+ca)<= a^2/ab + b^2/bc + c^2/ac
H.P.

Q2.

If P1, P2, ..., P2014 be an arbitrary rearrangement of 1, 2, ..., 2014.


Prove that

1/(P1 + P2) + 1/(P2 + P3) + ... + 1/(P2013 + P2014) > 2013/2016.

S2.
We will use Titu's lemma here again.
Which is:

a1^2/b1 + a2^2/b2 ... a2013^2/b2013 >= (a1 + a2 ... a2013)^2/(b1 + b2 ... b2013)

where ai are real, bi are positive.

Let all ai = 1 =>
1/b1 + 1/b2 .. 1/b2013 >= 2013^2/[2*(p2+p3...p2013) + p1 + p2014]

RHS denominator = 2*2014*2015/2 - p1 - p2014 = 2014*2015 - p1 - p2014

Now I want to show that LHS > 2013/2016
If I could show that RHS > 2013/2016 always, i.e. minimum value of RHS is > 2013/2016 then I am done.


To make the RHS smalles, denominator has to be largest.
So p1 + p2014 has to be smallest.
So p1+p2014 = 1 + 2

Denominator = 2014*2015 - 1 - 2 = 2014^2015 - 3
So RHS becomes:

2013^2/[2014^2015 - 3] > 2013/2016
=>
2013/[2014^2015 - 3] > 1/2016

Cross mult:
2016*2013 > 2014*2015 - 3

LHS:
(2015+1)(2014-1) = 2014*2015 - 1 + 1 > RHS
H.P.





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