Fie numerele reale x,y,z astfel incat \( x^2+y^2=z^2 \). Aflati maximul expresiei \( E=|\frac{x-y}{z}|(1+\frac{4xy}{z^2}) \).
G.M. 12/2008
Maximul expresiei
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BogdanCNFB
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\( x^2+y^2=z^2\Rightarrow (\frac{x}{z})^2+(\frac{y}{z})^2=1\Rightarrow\exists t\in\mathbb{R};\frac{x}{z}=\sin t,\frac{y}{z}=\cos t \).
Atunci avem \( E=|\sin t-\cos t|\cdot(1+4\sin t\cos t) \).
Notam \( \sin t-\cos t=k\Rightarrow \sin t\cdot\cos t=\frac{1-k^2}{2} \).
\( E=|k|\cdot(3-2k^2)\Rightarrow E^2=k^2(3-2k^2)^2\Rightarrow E^2=\frac{1}{4}\cdot 4k^2(3-2k^2)(3-2k^2)\le \frac{1}{4}\cdot[\frac{4k^2+(3-2k^2)+(3-2k^2)}{3}]^3=\frac{1}{4}\cdot 8=2\Rightarrow |E|\le \sqrt{2} \).
Deci maximul este \( \sqrt{2} \).
Atunci avem \( E=|\sin t-\cos t|\cdot(1+4\sin t\cos t) \).
Notam \( \sin t-\cos t=k\Rightarrow \sin t\cdot\cos t=\frac{1-k^2}{2} \).
\( E=|k|\cdot(3-2k^2)\Rightarrow E^2=k^2(3-2k^2)^2\Rightarrow E^2=\frac{1}{4}\cdot 4k^2(3-2k^2)(3-2k^2)\le \frac{1}{4}\cdot[\frac{4k^2+(3-2k^2)+(3-2k^2)}{3}]^3=\frac{1}{4}\cdot 8=2\Rightarrow |E|\le \sqrt{2} \).
Deci maximul este \( \sqrt{2} \).