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sửa đổi
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giuuuuuuuu
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giuuuuuuuu $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{(a+b+c)^ {2 }}{2\sqrt{3(ab+bc+ca)}}$
giuuuuuuuu $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{(a+b+c)^2}{2\sqrt{3(ab+bc+ca)}}$
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sửa đổi
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giuuuuuuuu
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giuuuuuuuu $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} $\geq \frac{(a+b+c)^2}{2\sqrt{3(ab+bc+ca)}} $
giuuuuuuuu $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{(a+b+c)^ {2 }}{2\sqrt{3(ab+bc+ca)}}
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sửa đổi
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giuuuuuuuu
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giuuuuuuuu $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{(a+b+c)^2}{2\sqrt{3(ab+bc+ca)}}$
giuuuuuuuu $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} $\geq \frac{(a+b+c)^2}{2\sqrt{3(ab+bc+ca)}}$
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sửa đổi
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giuuuuuuuu
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giuuuuuuuu $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$\geq \frac{(a+b+c)^2}{2\sqrt{3(ab+bc+ca)}}$
giuuuuuuuu $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$\geq \frac{(a+b+c)^2}{2\sqrt{3(ab+bc+ca)}}$
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sửa đổi
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ggggggggggggggggggg
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ggggggggggggggggggg cho a,b,c> tim min $\frac{b(a-c)}{c(a+b)}+\frac{c(3b+a)}{a(b+c)}+\frac{3c(a-b)}{b(a+c)}$
ggggggggggggggggggg cho a,b,c> 0 tim min $\frac{b(a-c)}{c(a+b)}+\frac{c(3b+a)}{a(b+c)}+\frac{3c(a-b)}{b(a+c)}$
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sửa đổi
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giiiiiiiiiiiup
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giiiiiiiiiiiup cho a,b,c>0 CMR:$\frac{b^2c^3}{a^2(b+c)^3}+\frac{c^2a^3}{b^2(c+a)^3}+\frac{a^2 c^3}{c^2(a+b)^3}\geq \frac{9abc}{4(3abc+ab^2+bc^2+c^2 a)}$
giiiiiiiiiiiup cho a,b,c>0 CMR:$\frac{b^2c^3}{a^2(b+c)^3}+\frac{c^2a^3}{b^2(c+a)^3}+\frac{a^2 b^3}{c^2(a+b)^3}\geq \frac{9abc}{4(3abc+ab^2+bc^2+c a^2)}$
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