Cho $a,b,c> 0, a+b+c=3$ Tìm GTNN
$P=\frac{a}{1+ab}+\frac{b}{1+bc}+\frac{c}{1+ac}$
Cho $a,b,c> 0, a+b+c=3$ Tìm GTNN
$P=\frac{a}{1+ab}+\frac{b}{1+bc}+\frac{c}{1+ac}$
Ta có: $\frac{a}{1+ab}=a-\frac{a^{2}b}{1+ab}\geq a-\frac{a^{2}b}{2\sqrt{ab}}=a-\frac{\sqrt{ab}}{2}$
Do đó: $P\geq a+b+c-\frac{1}{2}(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\geq a+b+c-\frac{1}{2}\frac{(a+b+c)^{2}}{3}=\frac{3}{2}$
Dấu = xảy ra <=> a = b = c = 1
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