1995 IMO Problems/Problem 2
Contents
[hide]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
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 5
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 6
We want to try and apply Cauchy (Titu), by transforming the numerator into a quadratic expression and the denominator into a linear expression. This is easily achieved by using the provided condition: \
Since . Likewise,
and
. Hence, by Cauchy (Titu):
\begin{align*}
\sum_{\text{cyc}}\frac{1}{a^3(b+c)} & \ge \frac{(bc+ac+ab)^2}{a(b+c)+b(a+c)+c(a+b)} \\& = \frac{(bc+ac+ab)^2}{2bc+2ac+2ab} \\& = \frac{bc+ac+ab}{2} \\& \ge \frac{3\sqrt[^3]{a^2b^2c^2}}{2}, \text{by AM-GM} \\& = \frac{3}{2}, \text{by the given condition}.
Solution 7 from Brilliant Wiki (Muirheads) =
https://brilliant.org/wiki/muirhead-inequality/
Scroll all the way down Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.