Difference between revisions of "2001 IMO Problems/Problem 2"
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but | but | ||
<cmath>1=(a+b+c)^3=a^3+b^3+c^3+6abc+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \geq a^3+b^3+c^3+24abc</cmath> by AM-GM, and thus the inequality is proven. | <cmath>1=(a+b+c)^3=a^3+b^3+c^3+6abc+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \geq a^3+b^3+c^3+24abc</cmath> by AM-GM, and thus the inequality is proven. | ||
+ | |||
+ | === Alternate Solution using Jensen's === | ||
+ | We can rewrite | ||
+ | <cmath>\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{c}{\sqrt{c^2+8ab}}</cmath> | ||
+ | as | ||
+ | <cmath>\frac{a}{\sqrt{a^2+\frac{8abc}{a}}}+\frac{b}{\sqrt{b^2+\frac{8abc}{b}}}+\frac{c}{\sqrt{c^2+\frac{8abc}{c}}}</cmath> | ||
+ | which is the same as | ||
+ | <cmath>\frac{\sqrt{a^3}}{\sqrt{a^3+8abc}}+\frac{\sqrt{b^3}}{\sqrt{b^3+8abc}}+\frac{\sqrt{c^3}}{\sqrt{c^3+8abc}}</cmath> | ||
+ | Now let <math>f(x)=\sqrt{\frac{x^3}{x^3 + abc}}</math>. Then f is concave, and f is strictly increasing, so by Jensen's inequality and AM-GM, | ||
+ | <cmath>f(a) + f(b) + f(c) \geq 3f((\frac{1}{3})a + (\frac{1}{3})b + (\frac{1}{3})c)) \geq 3f(\sqrt[3]{abc}) = 3(\frac{1}{3}) =1</cmath> | ||
=== Alternate Solution using Isolated Fudging === | === Alternate Solution using Isolated Fudging === |
Revision as of 15:37, 29 January 2019
Problem
Let be positive real numbers. Prove that
.
Contents
Solution
Firstly, (where
) and its cyclic variations.
Next note that
and
are similarly oriented sequences. Thus
Hence the inequality has been established.
Equality holds if
.
Notation: : AM-GM inequality,
: AM-HM inequality,
: Chebycheff's inequality,
: QM-AM inequality / RMS inequality
Alternate Solution using Holder's
By Holder's inequality,
Thus we need only show that
Which is obviously true since
.
Alternate Solution using Jensen's
This inequality is homogeneous so we can assume without loss of generality and apply Jensen's inequality for
, so we get:
but
by AM-GM, and thus the inequality is proven.
Alternate Solution using Jensen's
We can rewrite
as
which is the same as
Now let
. Then f is concave, and f is strictly increasing, so by Jensen's inequality and AM-GM,
Alternate Solution using Isolated Fudging
We claim that
Cross-multiplying, squaring both sides and expanding, we have
After cancelling the
term, we apply AM-GM to RHS and obtain
as desired, completing the proof of the claim.
Similarly and
.
Summing the three inequalities, we obtain the original inequality.
Alternate Solution using Cauchy
We want to prove
Note that since this inequality is homogenous, assume .
By Cauchy,
Dividing both sides by , we see that we want to prove
or equivalently
Squaring both sides, we have
Now use Cauchy again to obtain
Since , the inequality becomes
after some simplifying.
But this equals and since
we just want to prove
after some simplifying.
But that is true by AM-GM or Muirhead. Thus, proved.
See also
2001 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |