cho a,b,c>0 thỏa mãn $\sum a^{2}=1$. Tìm Min $P=\sum \frac{a^{2}}{b+c}$
Tìm Min $P=\sum \frac{a^{2}}{b+c}$
#1
Posted 23-11-2017 - 17:34
Knowing both victory and defeat.That is the way you become a real man-Shanks
#2
Posted 23-11-2017 - 17:58
cho a,b,c>0 thỏa mãn $\sum a^{2}=1$. Tìm Min $P=\sum \frac{a^{2}}{b+c}$
Ta có
$(a^{2}+b^{2}+c^{2})(a+b+c)=\sum a^{3}+\sum ab(a+b)\geq \sum \frac{a^{3}+b^{3}}{2}+\sum ab(a+b)\geq \frac{3}{2}\sum ab(a+b)\Rightarrow \sum ab(a+b)\leq \frac{2\sqrt{3}}{3}$
$P=\sum \frac{a^{2}}{b+c}=\sum \frac{a^{4}}{a^{2}(b+c)}\geq \frac{(\sum a^{2})^{2}}{\sum ab(a+b)}\geq \frac{\sqrt{3}}{2}$
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$\sqrt{VMF}$
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