Cho: $a,b,c,d> 0, c^{2}+d^{2}= (a^{2}+b^{2})^{3}. CMR \frac{a^{3}}{c}+\frac{b^{3}}{d}\geq 1$
Cho: $a,b,c,d> 0, c^{2}+d^{2}= (a^{2}+b^{2})^{3}. CMR \frac{a^{3}}{c}+\frac{b^{3}}{d}\geq 1$
Started By Hnim Naul, 31-07-2018 - 15:57
bất đẳng thức và cực trị
#1
Posted 31-07-2018 - 15:57
#2
Posted 31-07-2018 - 20:57
Cho: $a,b,c,d> 0, c^{2}+d^{2}= (a^{2}+b^{2})^{3}. CMR \frac{a^{3}}{c}+\frac{b^{3}}{d}\geq 1$
Áp dụng Cauchy-Schwarz, ta có
$VT\geq \frac{(a^{2}+b^{2})^{2}}{ac+bd}\geq \frac{(a^{2}+b^{2})^{2}}{\sqrt{(a^{2}+b^{2})(c^{2}+d^{2})}}=1$
- thanhdatqv2003 and ThinhThinh123 like this
#3
Posted 01-08-2018 - 08:39
[@bất đẳng thức Holder!]
$$\left ( \frac{a^{3}}{c}+ \frac{b^{3}}{d} \right )\left ( \frac{a^{3}}{c}+ \frac{b^{3}}{d} \right )\left ( c^{2}+ d^{2} \right )\geqq \left ( \sqrt[3]{\frac{a^{3}}{c}\,.\,\frac{a^{3}}{c}\,.\,c^{2}}+ \sqrt[3]{\frac{b^{3}}{c}\,.\,\frac{b^{3}}{c}\,.\,d^{2}} \right )^{3}= \left ( a^{2}+ b^{2} \right )^{3}$$
- thanhdatqv2003, ThinhThinh123 and Hnim Naul like this
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