Cho x, y ,z >0 va x + y + z <= 3/2
CMR. $\sqrt{x^2 + 1/x^2} + \sqrt{y^2 + 1/y^2} + \sqrt{z^2 + 1/z^2}>= (3/2) . $ $\sqrt{17}$
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Cho x, y ,z >0 va x + y + z <= 3/2
CMR. $\sqrt{x^2 + 1/x^2} + \sqrt{y^2 + 1/y^2} + \sqrt{z^2 + 1/z^2}>= (3/2) . $ $\sqrt{17}$
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$\displaystyle \begin{array}{{>{\displaystyle}l}} Theo\ Bunhiacopxki\ có:\\ \sqrt{x^{2} +\frac{1}{x^{2}}} =\frac{1}{\sqrt{1^{2} +4^{2}}}\sqrt{\left( x^{2} +\frac{1}{x^{2}}\right)\left( 1^{2} +4^{2}\right)} \geqslant \frac{1}{\sqrt{1^{2} +4^{2}}}\left( x+\frac{4}{x}\right)\\ \sqrt{y^{2} +\frac{1}{y^{2}}} =\frac{1}{\sqrt{1^{2} +4^{2}}}\sqrt{\left( y^{2} +\frac{1}{y^{2}}\right)\left( 1^{2} +4^{2}\right)} \geqslant \frac{1}{\sqrt{1^{2} +4^{2}}}\left( y+\frac{4}{y}\right)\\ \sqrt{z^{2} +\frac{1}{z^{2}}} =\frac{1}{\sqrt{1^{2} +4^{2}}}\sqrt{\left( z^{2} +\frac{1}{z^{2}}\right)\left( 1^{2} +4^{2}\right)} \geqslant \frac{1}{\sqrt{1^{2} +4^{2}}}\left( z+\frac{4}{z}\right)\\ \Rightarrow \sqrt{x^{2} +\frac{1}{x^{2}}} +\sqrt{y^{2} +\frac{1}{y^{2}}} +\sqrt{z^{2} +\frac{1}{z^{2}}} \geqslant \frac{1}{\sqrt{17}}\left[( x+y+z) +4\left(\frac{1}{x} +\frac{1}{y} +\frac{1}{z}\right)\right]\\ \geqslant \frac{1}{\sqrt{17}}\left[( x+y+z) +4\frac{9}{x+y+z}\right] \ \left( dễ\ thấy\ \frac{1}{x} +\frac{1}{y} +\frac{1}{z} \geqslant \frac{9}{x+y+z}\right)\\ =\frac{1}{\sqrt{17}}\left[( x+y+z) +\frac{9}{4}\frac{1}{x+y+z} +\frac{135}{4}\frac{1}{x+y+z}\right]\\ \geqslant \frac{1}{\sqrt{17}}\left( 2\sqrt{\frac{9}{4}} +\frac{135}{4}\frac{1}{x+y+z}\right) \geqslant \frac{1}{\sqrt{17}}\left( 2.\frac{3}{2} +\frac{2.135}{3.4}\right) =\frac{3\sqrt{17}}{2}\\ Dấu\ “=”\ xảy\ ra\ \Leftrightarrow x=y=z=\frac{1}{2} \end{array}$