Cho $0\leq x\leq 1$. CMR: $x(9\sqrt{1+x^2}+13\sqrt{1-x^2}) \leq 16$
$x(9\sqrt{1+x^2}+13\sqrt{1-x^2}) \leq 16$
Started By melodias2002, 01-02-2018 - 00:38
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Posted 01-02-2018 - 07:27
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Cho $0\leq x\leq 1$. CMR: $x(9\sqrt{1+x^2}+13\sqrt{1-x^2}) \leq 16$
$VT^{2}= x^{2}\left ( 13\sqrt{1- x^{2}} + 9\sqrt{1+ x^{2}}\right )= x^{2}\left ( \sqrt{13}. \sqrt{13- 13x^{2}}+ 3\sqrt{3}\sqrt{3+ 3x^{2}} \right )^{2}\leq 40x^{2}\left ( 16- 10x^{2} \right )= 4. \frac{\left ( 10x^{2}+ 16- 10x^{2} \right )^{2}}{4}= 16^{2}$
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