Difference between revisions of "1995 IMO Problems/Problem 2"
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=== Solution 7 from Brilliant Wiki (Muirheads) ==== | === Solution 7 from Brilliant Wiki (Muirheads) ==== | ||
https://brilliant.org/wiki/muirhead-inequality/ | https://brilliant.org/wiki/muirhead-inequality/ | ||
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+ | === Solution 8 (fast Titu's Lemma no substitutions) === | ||
+ | Rewrite <math>\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)}</math> as <math>\frac{(1/a)^2}{a(b+c)} + \frac{(1/b)^2}{b(a+c)} + \frac{(1/c)^2}{c(a+b)}</math>. | ||
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+ | Now applying Titu's lemma yields <math>\frac{(1/a)^2}{a(b+c)} + \frac{(1/b)^2}{b(a+c)} + \frac{(1/c)^2}{c(a+b)} \geq \frac{(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2}{a(b+c) + b(a+c) + c(a+b)} = \frac{(ab + bc + ca)^2}{2(ab + bc + ca)} = \frac{ab + bc + ca}{2}</math>. | ||
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+ | Now applying the AM-GM inequality on <math>ab + bc +ca \geq 3((abc)^2)^{\frac{1}{3}} = 3</math>. The result now follows. | ||
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+ | Note: <math>ab + bc + ca = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}</math>, because <math>abc = 1</math>. (Why? Because <math>a = \frac{1}{bc}</math>, and hence <math>\frac{1}{a} = bc</math>). | ||
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+ | ~th1nq3r | ||
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{{alternate solutions}} | {{alternate solutions}} | ||
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== Resources == | == Resources == |
Revision as of 12:23, 6 January 2022
Contents
Problem
(Nazar Agakhanov, Russia)
Let be positive real numbers such that
. Prove that
Solution
Solution 1
We make the substitution ,
,
. Then
Since
and
are similarly sorted sequences, it follows from the Rearrangement Inequality that
By the Power Mean Inequality,
Symmetric application of this argument yields
Finally, AM-GM gives us
as desired.
Solution 2
We make the same substitution as in the first solution. We note that in general,
It follows that
and
are similarly sorted sequences. Then by Chebyshev's Inequality,
By AM-GM,
, and by Nesbitt's Inequality,
The desired conclusion follows.
Solution 3
Without clever substitutions:
By Cauchy-Schwarz, Dividing by
gives
by AM-GM.
Solution 3b
Without clever notation:
By Cauchy-Schwarz,
Dividing by and noting that
by AM-GM gives
as desired.
Solution 4
After the setting and as
so
concluding
By Titu Lemma,
Now by AM-GM we know that
and
which concludes to
Therefore we get
Hence our claim is proved ~~ Aritra12
Solution 5
Proceed as in Solution 1, to arrive at the equivalent inequality
But we know that
by AM-GM. Furthermore,
by Cauchy-Schwarz, and so dividing by
gives
as desired.
Solution 6
Without clever substitutions, and only AM-GM!
Note that . The cyclic sum becomes
. Note that by AM-GM, the cyclic sum is greater than or equal to
. We now see that we have the three so we must be on the right path. We now only need to show that
. Notice that by AM-GM,
,
, and
. Thus, we see that
, concluding that
Solution 7 from Brilliant Wiki (Muirheads) =
https://brilliant.org/wiki/muirhead-inequality/
Solution 8 (fast Titu's Lemma no substitutions)
Rewrite as
.
Now applying Titu's lemma yields .
Now applying the AM-GM inequality on . The result now follows.
Note: , because
. (Why? Because
, and hence
).
~th1nq3r
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Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.