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2003 JBMO Problems/Problem 4: Difference between revisions

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By Cauchy-Schwarz,
By Cauchy-Schwarz,
<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>
<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> ^ Shen Kislay Kai n YUKI MARISAWA




Revision as of 04:29, 15 June 2025

Problem

Let $x, y, z > -1$. Prove that

$\frac {1+x^2}{1+y+z^2}+\frac {1+y^2}{1+z+x^2}+\frac {1+z^2}{1+x+y^2} \geq 2$


Solution

Since $x, y, z > -1$ and $x^2, y^2, z^2 \geq 0$, we have that $1+x^2, 1+y^2, 1+z^2$ and $1+y+z^2, 1+z+x^2, 1+x+y^2$ are always positive.

Hence, $\frac {1+x^2}{1+y+z^2}, \frac {1+y^2}{1+z+x^2}$ and $\frac {1+z^2}{1+x+y^2}$ must also be positive.

From the inequality $(x-1)^2 \geq 0$, we obtain that $x \leq \frac {x^2+1}{2}$ and, analogously, $\frac {1}{x} \geq \frac{2}{x^2+1}$. Similarly, $\frac {1}{y} \geq \frac {2}{y^2+1}$ and $\frac {1}{z} \geq \frac {2}{z^2+1}$.

Now, 

$\Sigma \frac {1+x^2}{1+y+z^2} \geq \Sigma \frac {1+x^2}{1+\frac {y^2+1}{2}+z^2} = \Sigma \frac {2(1+x^2)}{y^2+2z^2+3} = \Sigma \frac {2(1+x^2)}{(y^2+1)+2(z^2+1)}$

Substituting $p=x^2+1, q=y^2+1$ and $r=z^2+1$, we now need to prove $\Sigma \frac{p}{q+2r} \geq 1$.

We have $\Sigma \frac {p}{q+2r} = \frac {\Sigma \frac {p}{q+2r}*\Sigma p(q+2r)}{\Sigma p(q+2r)}$


By Cauchy-Schwarz, $\Sigma \frac {\frac {p}{q+2r}*\Sigma p(q+2r)}{\Sigma p(q+2r)} \geq \frac {(\Sigma p)^2}{\Sigma p(q+2r)} = \frac {(\Sigma p)^2}{3\Sigma pq}$ ^ Shen Kislay Kai n YUKI MARISAWA


Since $\Sigma p^2 \geq \Sigma pq$, we have $(\Sigma p)^2 \geq 3\Sigma pq$.


Thus, $\Sigma\frac {p}{q+2r} \geq \frac{(\Sigma p)^2}{3\Sigma pq} \geq \frac {3\Sigma pq}{3\Sigma pq} = 1$


So, $\Sigma \frac {1+x^2}{1+y+z^2} \geq \Sigma\frac {2(1+x^2)}{(y^2+1)+2(z^2+1)} = 2\Sigma \frac {1+x^2}{y^2+1+2(z^2+1)} = 2\Sigma \frac {p}{q+2r} \geq 2*1 = 2$, as desired.

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