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} | + | 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>\left(\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1} {c}=1\right)</math> concluding <math>x y z=1 .</math> |
<math>\textsf{Claim}:</math> | <math>\textsf{Claim}:</math> | ||
<cmath> | <cmath> | ||
− | \frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{3}{2 | + | \frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{3}{2} |
</cmath> | </cmath> | ||
By Titu Lemma, | By Titu Lemma, | ||
Line 63: | Line 63: | ||
</cmath> | </cmath> | ||
Now by AM-GM we know that <cmath> (x+y+z)\geq3\sqrt[3]{xyz} | Now by AM-GM we know that <cmath> (x+y+z)\geq3\sqrt[3]{xyz} | ||
− | </cmath>and <math>xyz=1</math> which concludes to <math> | + | </cmath>and <math>xyz=1</math> which concludes to <math>\implies (x+y+z)\geq3\sqrt[3]{1}</math> |
Therefore we get | Therefore we get |
Revision as of 09:42, 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
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/
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.