Difference between revisions of "1995 IMO Problems/Problem 2"
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=== Solution 4 === | === Solution 4 === | ||
+ | After the setting <math>a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z},</math> and as <math>abc=1</math> so <math>\frac{1}{a}.\frac{1}{b}.\frac{1} {c}=1</math> concluding <math>x y z=1 .</math> | ||
+ | |||
+ | |||
+ | \[ | ||
+ | \boxed{\textbf{Claim}:\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{3}{2}} | ||
+ | \] | ||
+ | By Titu Lemma, | ||
+ | \[ | ||
+ | \implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{(x+y+z)^{2}}{2(x+y+z)} | ||
+ | \] | ||
+ | \[ | ||
+ | \implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{(x+y+z)}{2} | ||
+ | \] | ||
+ | Now by AM-GM we know that\[ (x+y+z)\geq3\sqrt[3]{xyz} | ||
+ | \]and <math>xyz=1</math> which concludes to <math>\implies (x+y+z)\geq3\sqrt[3]{1}</math> | ||
+ | |||
+ | Therefore we get | ||
+ | |||
+ | \[ | ||
+ | \implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{3}{2} | ||
+ | \]Hence our claim is proved ~~ Aritra12 | ||
+ | |||
+ | === Solution 5 === | ||
Proceed as in Solution 1, to arrive at the equivalent inequality | Proceed as in Solution 1, to arrive at the equivalent inequality | ||
<cmath> \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{3}{2} . </cmath> | <cmath> \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{3}{2} . </cmath> | ||
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as desired. | as desired. | ||
− | === Solution | + | === Solution 6 === |
Without clever substitutions, and only AM-GM! | Without clever substitutions, and only AM-GM! | ||
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− | === Solution | + | === Solution 7 from Brilliant Wiki (Muirheads) ==== |
https://brilliant.org/wiki/muirhead-inequality/ | https://brilliant.org/wiki/muirhead-inequality/ | ||
Revision as of 09:38, 30 January 2021
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
After the setting and as
so
concluding
\[
\boxed{\textbf{Claim}:\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{3}{2}}
\]
By Titu Lemma,
\[
\implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{(x+y+z)^{2}}{2(x+y+z)}
\]
\[
\implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{(x+y+z)}{2}
\]
Now by AM-GM we know that\[ (x+y+z)\geq3\sqrt[3]{xyz}
\]and which concludes to
Therefore we get
\[ \implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{3}{2} \]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/
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.