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bình luận
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Tìm GTLN
doan la a=b=0;c=1.hok biet dung hay sai
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bình luận
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Tìm GTLN
de dung ma,hok sai dau
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đặt câu hỏi
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chung minh bdt
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$a,b,c>0;abc=1$.chung minh
$\frac{a}{1+c+b} + \frac{b}{1+c+a} + \frac{c}{1+a+b} \geqslant 1$
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sửa đổi
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bat dang thuc
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dat A=\frac{a^{2}}{ba+ca}+\frac{b^{2}}{bc+ab}+\frac{c^{2}}{bc+ac}\Rightarrow A\times \left ( ab+bc+ca \right )\geqslant \left ( a+b+c \right )^{2}\Rightarrow A\geqslant \frac{a^{2}+b^{2}+c^{2}+2ab+2bc+2ca}{2ab+2bc+2ca}\geqslant \frac{3}{2}vi:a^{2}+b^{2}+c^{2}\geqslant ab+bc+ca
A=\frac{a^{2}}{ba+ca}+\frac{b^{2}}{bc+ab}+\frac{c^{2}}{bc+ac}\Rightarrow A\times \left ( ab+bc+ca \right )\geqslant \left ( a+b+c \right )^{2}\Rightarrow A\geqslant \frac{a^{2}+b^{2}+c^{2}+2ab+2bc+2ca}{2ab+2bc+2ca}\geqslant \frac{3}{2}vi:a^{2}+b^{2}+c^{2}\geqslant ab+bc+ca
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đặt câu hỏi
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cho:a+b+c=1,tim Max
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Cho $a,b,c >0,a+b+c=1$. Tìm GTLN :
$P=\frac{1+a^{2}}{1+b^{2}}+\frac{1+b^{2}}{1+c^{2}}+\frac{1+c^{2}}{1+a^{2}}$
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giải đáp
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bat dang thuc
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$A=\frac{a^{2}}{ba+ca}+\frac{b^{2}}{bc+ab}+\frac{c^{2}}{bc+ac}$ $\Rightarrow A\times \left ( ab+bc+ca \right )\geqslant \left ( a+b+c \right )^{2}$
$\Rightarrow A\geqslant \frac{a^{2}+b^{2}+c^{2}+2ab+2bc+2ca}{2ab+2bc+2ca}\geqslant \frac{3}{2}$
vi:$a^{2}+b^{2}+c^{2}\geqslant ab+bc+ca$ |
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