Difference between revisions of "2003 JBMO Problems/Problem 4"
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We have | We have | ||
<math>\Sigma \frac {p}{q+2r} = \frac {\Sigma \frac {p}{q+2r}*\Sigma p(q+2r)}{\Sigma p(q+2r)}</math> | <math>\Sigma \frac {p}{q+2r} = \frac {\Sigma \frac {p}{q+2r}*\Sigma p(q+2r)}{\Sigma p(q+2r)}</math> | ||
+ | |||
By Cauchy-Schwarz, | By Cauchy-Schwarz, | ||
− | <math>\Sigma \frac {\frac {p}{q+2r}*\Sigma p | + | <math>\Sigma \frac {\frac {p}{q+2r}*\Sigma p(q+2r)}{\Sigma p(q+2r)} \geq \frac {(\Sigma p)^2}{\Sigma p(q+2r)} = \frac {(\Sigma p)^2}{3\Sigma pq}</math> |
+ | |||
Since <math>\Sigma p^2 \geq \Sigma pq</math>, we have <math>(\Sigma p)^2 \geq 3\Sigma pq</math>. | Since <math>\Sigma p^2 \geq \Sigma pq</math>, we have <math>(\Sigma p)^2 \geq 3\Sigma pq</math>. | ||
+ | |||
Thus, | Thus, | ||
<math>\Sigma\frac {p}{q+2r} \geq \frac{(\Sigma p)^2}{3\Sigma pq} \geq \frac {3\Sigma pq}{3\Sigma pq} = 1</math> | <math>\Sigma\frac {p}{q+2r} \geq \frac{(\Sigma p)^2}{3\Sigma pq} \geq \frac {3\Sigma pq}{3\Sigma pq} = 1</math> | ||
+ | |||
So, | So, | ||
− | <math>\Sigma \frac {1+x^2}{1+y+z^2} \geq \Sigma\frac {2(1+x^2)}{(y^2+1)+2(z^2+1)} = 2\Sigma \frac {1+x^2}{y^2+1+2(z^2+1)} = 2\Sigma \frac {p}{q+2r} \geq 2*1 = 2</math> | + | <math>\Sigma \frac {1+x^2}{1+y+z^2} \geq \Sigma\frac {2(1+x^2)}{(y^2+1)+2(z^2+1)} = 2\Sigma \frac {1+x^2}{y^2+1+2(z^2+1)} = 2\Sigma \frac {p}{q+2r} \geq 2*1 = 2</math>, as desired. |
Latest revision as of 18:17, 25 April 2021
Problem
Let . Prove that
Solution
Since and
, we have that
and
are always positive.
Hence, and
must also be positive.
From the inequality , we obtain that
and, analogously,
. Similarly,
and
.
Now,
Substituting and
, we now need to prove
.
We have
By Cauchy-Schwarz,
Since , we have
.
Thus,
So,
, as desired.