9k ill try
i personally lost interest in solving thes types of inequalities after i found that theorem . why try doin the hard way using proofs that cauchy swarchz or wathever others used wen u have law of symmetry
try to prove the following inequality
given that a,b,c are positive reals
Cyclic[(a^2) / {b(a + 2b)}] >= 1
a harddddddddddddddddd problem
not at all easy unless u guys know Cauchy Schwarz inequality
-
UP 0 DOWN 0 0 15
15 Answers
9yes i know
i stated a theorem which i found wile prep for rmo
when i posted it it got deleted
ill repost
it states that for any cyclic func max or min occurs wen all parameters are equal. it follows from symettry and i ve a big logical proof for that
im not letting it out till i do a phd on that;)
9
341I have a bit of a convoluted proof:
We have to prove that Σcyc a2/(ab+2b2) ≥1
If we let x = b/a, y = c/b and z = a/c, the inequality becomes:
Σcyc 1/(2x2+2x) ≥ 1 with xyz=1
Now for one more round of substitution,
Let x = pq/r2, y = qr/p2, z = rp/q2
Then we have to prove that
Σcyc r4/(2p2q2+pqr2) ≥1
This is where Cauchy Schwarz comes in:
Σcyc r4/(2p2q2+pqr2) ≥ (p2+q2+r2)2 /(2Σp2q2+Σpqr2) from Cauchy Schwarz ( in the form popularly known as Titu's lemma)
Now, if we prove that (p2+q2+r2)2 /(2Σp2q2+Σpqr2) ≥1, we are through
This is equivalent to proving that
(p2+q2+r2)2 ≥2Σp2q2+Σpqr2
or
Σp4 + 2Σp2q2 ≥2Σp2q2+Σpqr2
or Σp4 ≥ Σpqr2
which is true by Rearrangement Inequality or Muirhead as (4,0,0) majorizes (2,1,1)
341Of course, the last inequality has a solution by AM-GM too (as indicated by a Muirhead solution)
We have the following inequalities by AM-GM,
p4+q4+r4+r4 ≥ 4 pqr2.........1
p4+q4+q4+r4 ≥ 4 pq2r..........2
p4+p4+q4+r4 ≥ 4p2qr...........3
Adding 1,2 and 3, we get
4(p4+q4+r4) ≥ 4(pqr2+pq2r+p2qr) or
Σp4 ≥ Σpqr2 which is the required inequality
341See, from Cauchy Schwarz we get
(x1+x2+....+xn) (a12/x1+a22/x2+...+an2/xn) ≥ (a1+a2+...+an)2
The same is written in a more amenable form as:
a12/x1+a22/x2+...+an2/xn ≥ (a1+a2+...+an)2/(x1+x2+...+xn)
where the xi's are all positive
This is known as Titu's Lemma as it was popularised by Titu Andreescu.
1ok i think i have a better proof
you see lets take the first term
i.e
a2/b(a+2b)
=>
1/((b/a)+2(b/a)2)
take b/a=A c/b=B , a/c=C
then we need to prove
Σ1/2A2+A>=1
with ABC=1
now put 1/A=x,1/B=y,1/C=z
then we get to prove
Σx2/(x+2)>=1
with xyz=1;
byt titu's lemma
the above >= (x+y+z)2/((x+y+z)+6)
but as x+y+z>3(xyz)1/3>=3 (by A.M>G.M)
let x+y+z=t
we need to prove
t2 - t - 6>=0
that is proved as t>=3..